Please help me understand the following solution












3












$begingroup$


Prove by induction of n



$sumlimits_{k=1}^n frac k{k+1} leq n - frac1{n+1}$





begin{align}sum_1^{n+1}frac k{k+1}&leq n-frac 1{n+1}+frac{n+1}{n+2}\&=n-frac 1{n+1}+1-frac 1{n+2}\&=(n+1)-frac{2(n+2)-1}{(n+1)(n+2)}\&=(n+1)-frac 2{n+1}+frac 1{(n+1)(n+2)}\&leq (n+1)-frac 2{n+2}+frac 1{n+2}=(n+1)-frac 1{n+2}end{align}





Now I'm a beginner at induction, and couldn't follow this solution very well.I was hoping someone could help break down the steps and explain them.



Questions




  1. How the inequality works


Wouldn't



$sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1} $



become



$sum_1^{n+1}frac k{k+1} +frac{n+1}{n+2} leq n-frac 1{n+1}$



and then



$sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1}-frac{n+1}{n+2} $



instead of



$sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1}+frac{n+1}{n+2} $






  1. My largest issue with induction, is when the inequalities change like in the first and last step. I don't understand how that works. Any explanation, or good resources to help with my understanding of how the inequality changes when performing induction would be helpful.










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$endgroup$

















    3












    $begingroup$


    Prove by induction of n



    $sumlimits_{k=1}^n frac k{k+1} leq n - frac1{n+1}$





    begin{align}sum_1^{n+1}frac k{k+1}&leq n-frac 1{n+1}+frac{n+1}{n+2}\&=n-frac 1{n+1}+1-frac 1{n+2}\&=(n+1)-frac{2(n+2)-1}{(n+1)(n+2)}\&=(n+1)-frac 2{n+1}+frac 1{(n+1)(n+2)}\&leq (n+1)-frac 2{n+2}+frac 1{n+2}=(n+1)-frac 1{n+2}end{align}





    Now I'm a beginner at induction, and couldn't follow this solution very well.I was hoping someone could help break down the steps and explain them.



    Questions




    1. How the inequality works


    Wouldn't



    $sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1} $



    become



    $sum_1^{n+1}frac k{k+1} +frac{n+1}{n+2} leq n-frac 1{n+1}$



    and then



    $sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1}-frac{n+1}{n+2} $



    instead of



    $sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1}+frac{n+1}{n+2} $






    1. My largest issue with induction, is when the inequalities change like in the first and last step. I don't understand how that works. Any explanation, or good resources to help with my understanding of how the inequality changes when performing induction would be helpful.










    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Prove by induction of n



      $sumlimits_{k=1}^n frac k{k+1} leq n - frac1{n+1}$





      begin{align}sum_1^{n+1}frac k{k+1}&leq n-frac 1{n+1}+frac{n+1}{n+2}\&=n-frac 1{n+1}+1-frac 1{n+2}\&=(n+1)-frac{2(n+2)-1}{(n+1)(n+2)}\&=(n+1)-frac 2{n+1}+frac 1{(n+1)(n+2)}\&leq (n+1)-frac 2{n+2}+frac 1{n+2}=(n+1)-frac 1{n+2}end{align}





      Now I'm a beginner at induction, and couldn't follow this solution very well.I was hoping someone could help break down the steps and explain them.



      Questions




      1. How the inequality works


      Wouldn't



      $sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1} $



      become



      $sum_1^{n+1}frac k{k+1} +frac{n+1}{n+2} leq n-frac 1{n+1}$



      and then



      $sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1}-frac{n+1}{n+2} $



      instead of



      $sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1}+frac{n+1}{n+2} $






      1. My largest issue with induction, is when the inequalities change like in the first and last step. I don't understand how that works. Any explanation, or good resources to help with my understanding of how the inequality changes when performing induction would be helpful.










      share|cite|improve this question









      $endgroup$




      Prove by induction of n



      $sumlimits_{k=1}^n frac k{k+1} leq n - frac1{n+1}$





      begin{align}sum_1^{n+1}frac k{k+1}&leq n-frac 1{n+1}+frac{n+1}{n+2}\&=n-frac 1{n+1}+1-frac 1{n+2}\&=(n+1)-frac{2(n+2)-1}{(n+1)(n+2)}\&=(n+1)-frac 2{n+1}+frac 1{(n+1)(n+2)}\&leq (n+1)-frac 2{n+2}+frac 1{n+2}=(n+1)-frac 1{n+2}end{align}





      Now I'm a beginner at induction, and couldn't follow this solution very well.I was hoping someone could help break down the steps and explain them.



      Questions




      1. How the inequality works


      Wouldn't



      $sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1} $



      become



      $sum_1^{n+1}frac k{k+1} +frac{n+1}{n+2} leq n-frac 1{n+1}$



      and then



      $sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1}-frac{n+1}{n+2} $



      instead of



      $sum_1^{n+1}frac k{k+1}leq n-frac 1{n+1}+frac{n+1}{n+2} $






      1. My largest issue with induction, is when the inequalities change like in the first and last step. I don't understand how that works. Any explanation, or good resources to help with my understanding of how the inequality changes when performing induction would be helpful.







      discrete-mathematics induction






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      asked 2 hours ago









      BrownieBrownie

      997




      997






















          3 Answers
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          4












          $begingroup$

          Remember that in induction proofs, we start by assuming that the claim we're trying to prove is true for $n$, and then conclude that it must also be true for $n + 1$. In this case, we start with the induction assumption
          $$sum_{k = 1}^{n} frac{k}{k + 1} leq n - frac{1}{n + 1},$$
          and want to end with the conclusion that $$sum_{k = 1}^{n + 1} frac{k}{k + 1} leq n + 1 - frac{1}{n + 2} .$$
          Here's the argument written out in a bit more detail with commentary on each step.
          begin{align*}
          sum_{k = 1}^{n + 1} frac{k}{k + 1} & = frac{n + 1}{n + 2} + sum_{k = 1}^{n} frac{k}{k + 1} & textrm{ (just writing out the sum)} \
          & leq frac{n + 1}{n + 2} + n - frac{1}{n + 1} & textrm{ (applying the induction hypothesis)} \
          & = 1 - frac{1}{n + 2} + n - frac{1}{n + 1} & textrm{ (rewriting $frac{n+ 1}{n + 2}$ as $frac{n + 2 - 1}{n + 2} = 1 - frac{1}{n + 2}$)} \
          & = n + 1 - frac{1}{n + 1} - frac{1}{n + 2} & textrm{ (regrouping)} \
          & = n + 1 - frac{(n + 2) + (n + 1)}{(n + 2)(n + 1)} & textrm{ (combining fractions)} \
          & = n + 1 - frac{2(n + 2) - 1}{(n + 1)(n + 2)} & textrm{ (regrouping the numerator)} \
          & = n + 1 - frac{2(n + 2)}{(n + 1)(n + 2)} + frac{1}{(n + 1)(n + 2)} & textrm{ (breaking the fraction back apart)} \
          & = n + 1 - frac{2}{n + 1} + frac{1}{(n + 1)(n + 2)} & textrm{ (simplifying the fraction)} \
          & leq n + 1 - frac{2}{n + 2} + frac{1}{(n + 1)(n + 2)} & textrm{ (we slightly modified the second-to-last summand)} \
          & leq n + 1 - frac{2}{n + 2} + frac{1}{n + 2} & textrm{ (modifying the last summand)} \
          & = n + 1 - frac{1}{n + 2} & textrm{ (combining the fractions)} .
          end{align*}

          So in summary, we began with the assumption that the claim held for $n$, and through some arithmetic trickery concluded that it therefore held for $n + 1$.






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          $endgroup$









          • 1




            $begingroup$
            This is an amazing break down, could explain your second-to-last and third-to-last steps where you modify the summand? Oh I think I might understand, is it to make the inequality <= ?
            $endgroup$
            – Brownie
            1 hour ago








          • 1




            $begingroup$
            @Brownie In the third-to-last step, we observe that $n + 1 leq n + 2$, so $frac{2}{n + 1} geq frac{2}{n + 2}$, so $- frac{2}{n + 1} leq - frac{2}{n + 2}$. For the second-to-last step, we can see that $frac{1}{n + 2} = frac{n + 1}{(n + 1)(n + 2)} = (n + 1) frac{1}{(n + 1)(n + 2)} geq frac{1}{(n + 1)(n + 2)}$.
            $endgroup$
            – AJY
            1 hour ago








          • 1




            $begingroup$
            So is this a step you saw you could implement to get the final equal to $ n + 1 - frac{1}{n + 2} $ ? Or is there something that would push you towards doing this?
            $endgroup$
            – Brownie
            1 hour ago








          • 2




            $begingroup$
            @Brownie The short answer is that these kinds of tricks come easier with practice. We wanted to get to a very particular estimate, and just kinda made what estimates we could until it fell out just right. Sometimes (often) it just takes trial and error.
            $endgroup$
            – AJY
            1 hour ago



















          1












          $begingroup$

          Regarding the first inequality you're asking about, they are actually adding the term $frac{n+1}{n+2}$ to both sides, but on the left hand side it is added by increasing the upper limit in the sum by $1$, and to not change the inequality you have to add $frac{n+1}{n+2}$ to the right hand side.



          The subsequent steps in the proof you posted are just rearrangements of fractions using algebra, nothing actually changes there until the last line. Then they just use the fact that $frac{1}{(n+1)(n+2)} leq frac{1}{n+2}$






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          $endgroup$













          • $begingroup$
            Cool just last quick question, for the left hand side if instead of increasing the upper limit of the sum by 1, I added $frac{n+1}{n+2}$ to both sides, that would be the same?
            $endgroup$
            – Brownie
            2 hours ago










          • $begingroup$
            Yes, that would be the same.
            $endgroup$
            – Thomas Fjærvik
            2 hours ago



















          1












          $begingroup$

          If $aleq b $, then $a+cleq b+c $ for any $cin Bbb R $. By assumption, we have $$sum_{k=1}^{n}frac k{k+1}leq n-frac 1{n+1}.$$ Now add $dfrac{n+1}{n+2}$ on both sides, i.e., $$sum_{k=1}^{n}frac k{k+1}+frac{n+1}{n+2}leq n-frac 1{n+1}+frac{n+1}{n+2}.$$ Note that $$sum_{k=1}^{n}frac k{k+1}+frac{n+1}{n+2}=sum_{k=1}^{n+1}frac k{k+1}.$$ Hence we have $$sum_{k=1}^{n+1}frac k{k+1}leq n-frac 1{n+1}+frac{n+1}{n+2}.$$






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            $begingroup$

            Remember that in induction proofs, we start by assuming that the claim we're trying to prove is true for $n$, and then conclude that it must also be true for $n + 1$. In this case, we start with the induction assumption
            $$sum_{k = 1}^{n} frac{k}{k + 1} leq n - frac{1}{n + 1},$$
            and want to end with the conclusion that $$sum_{k = 1}^{n + 1} frac{k}{k + 1} leq n + 1 - frac{1}{n + 2} .$$
            Here's the argument written out in a bit more detail with commentary on each step.
            begin{align*}
            sum_{k = 1}^{n + 1} frac{k}{k + 1} & = frac{n + 1}{n + 2} + sum_{k = 1}^{n} frac{k}{k + 1} & textrm{ (just writing out the sum)} \
            & leq frac{n + 1}{n + 2} + n - frac{1}{n + 1} & textrm{ (applying the induction hypothesis)} \
            & = 1 - frac{1}{n + 2} + n - frac{1}{n + 1} & textrm{ (rewriting $frac{n+ 1}{n + 2}$ as $frac{n + 2 - 1}{n + 2} = 1 - frac{1}{n + 2}$)} \
            & = n + 1 - frac{1}{n + 1} - frac{1}{n + 2} & textrm{ (regrouping)} \
            & = n + 1 - frac{(n + 2) + (n + 1)}{(n + 2)(n + 1)} & textrm{ (combining fractions)} \
            & = n + 1 - frac{2(n + 2) - 1}{(n + 1)(n + 2)} & textrm{ (regrouping the numerator)} \
            & = n + 1 - frac{2(n + 2)}{(n + 1)(n + 2)} + frac{1}{(n + 1)(n + 2)} & textrm{ (breaking the fraction back apart)} \
            & = n + 1 - frac{2}{n + 1} + frac{1}{(n + 1)(n + 2)} & textrm{ (simplifying the fraction)} \
            & leq n + 1 - frac{2}{n + 2} + frac{1}{(n + 1)(n + 2)} & textrm{ (we slightly modified the second-to-last summand)} \
            & leq n + 1 - frac{2}{n + 2} + frac{1}{n + 2} & textrm{ (modifying the last summand)} \
            & = n + 1 - frac{1}{n + 2} & textrm{ (combining the fractions)} .
            end{align*}

            So in summary, we began with the assumption that the claim held for $n$, and through some arithmetic trickery concluded that it therefore held for $n + 1$.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              This is an amazing break down, could explain your second-to-last and third-to-last steps where you modify the summand? Oh I think I might understand, is it to make the inequality <= ?
              $endgroup$
              – Brownie
              1 hour ago








            • 1




              $begingroup$
              @Brownie In the third-to-last step, we observe that $n + 1 leq n + 2$, so $frac{2}{n + 1} geq frac{2}{n + 2}$, so $- frac{2}{n + 1} leq - frac{2}{n + 2}$. For the second-to-last step, we can see that $frac{1}{n + 2} = frac{n + 1}{(n + 1)(n + 2)} = (n + 1) frac{1}{(n + 1)(n + 2)} geq frac{1}{(n + 1)(n + 2)}$.
              $endgroup$
              – AJY
              1 hour ago








            • 1




              $begingroup$
              So is this a step you saw you could implement to get the final equal to $ n + 1 - frac{1}{n + 2} $ ? Or is there something that would push you towards doing this?
              $endgroup$
              – Brownie
              1 hour ago








            • 2




              $begingroup$
              @Brownie The short answer is that these kinds of tricks come easier with practice. We wanted to get to a very particular estimate, and just kinda made what estimates we could until it fell out just right. Sometimes (often) it just takes trial and error.
              $endgroup$
              – AJY
              1 hour ago
















            4












            $begingroup$

            Remember that in induction proofs, we start by assuming that the claim we're trying to prove is true for $n$, and then conclude that it must also be true for $n + 1$. In this case, we start with the induction assumption
            $$sum_{k = 1}^{n} frac{k}{k + 1} leq n - frac{1}{n + 1},$$
            and want to end with the conclusion that $$sum_{k = 1}^{n + 1} frac{k}{k + 1} leq n + 1 - frac{1}{n + 2} .$$
            Here's the argument written out in a bit more detail with commentary on each step.
            begin{align*}
            sum_{k = 1}^{n + 1} frac{k}{k + 1} & = frac{n + 1}{n + 2} + sum_{k = 1}^{n} frac{k}{k + 1} & textrm{ (just writing out the sum)} \
            & leq frac{n + 1}{n + 2} + n - frac{1}{n + 1} & textrm{ (applying the induction hypothesis)} \
            & = 1 - frac{1}{n + 2} + n - frac{1}{n + 1} & textrm{ (rewriting $frac{n+ 1}{n + 2}$ as $frac{n + 2 - 1}{n + 2} = 1 - frac{1}{n + 2}$)} \
            & = n + 1 - frac{1}{n + 1} - frac{1}{n + 2} & textrm{ (regrouping)} \
            & = n + 1 - frac{(n + 2) + (n + 1)}{(n + 2)(n + 1)} & textrm{ (combining fractions)} \
            & = n + 1 - frac{2(n + 2) - 1}{(n + 1)(n + 2)} & textrm{ (regrouping the numerator)} \
            & = n + 1 - frac{2(n + 2)}{(n + 1)(n + 2)} + frac{1}{(n + 1)(n + 2)} & textrm{ (breaking the fraction back apart)} \
            & = n + 1 - frac{2}{n + 1} + frac{1}{(n + 1)(n + 2)} & textrm{ (simplifying the fraction)} \
            & leq n + 1 - frac{2}{n + 2} + frac{1}{(n + 1)(n + 2)} & textrm{ (we slightly modified the second-to-last summand)} \
            & leq n + 1 - frac{2}{n + 2} + frac{1}{n + 2} & textrm{ (modifying the last summand)} \
            & = n + 1 - frac{1}{n + 2} & textrm{ (combining the fractions)} .
            end{align*}

            So in summary, we began with the assumption that the claim held for $n$, and through some arithmetic trickery concluded that it therefore held for $n + 1$.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              This is an amazing break down, could explain your second-to-last and third-to-last steps where you modify the summand? Oh I think I might understand, is it to make the inequality <= ?
              $endgroup$
              – Brownie
              1 hour ago








            • 1




              $begingroup$
              @Brownie In the third-to-last step, we observe that $n + 1 leq n + 2$, so $frac{2}{n + 1} geq frac{2}{n + 2}$, so $- frac{2}{n + 1} leq - frac{2}{n + 2}$. For the second-to-last step, we can see that $frac{1}{n + 2} = frac{n + 1}{(n + 1)(n + 2)} = (n + 1) frac{1}{(n + 1)(n + 2)} geq frac{1}{(n + 1)(n + 2)}$.
              $endgroup$
              – AJY
              1 hour ago








            • 1




              $begingroup$
              So is this a step you saw you could implement to get the final equal to $ n + 1 - frac{1}{n + 2} $ ? Or is there something that would push you towards doing this?
              $endgroup$
              – Brownie
              1 hour ago








            • 2




              $begingroup$
              @Brownie The short answer is that these kinds of tricks come easier with practice. We wanted to get to a very particular estimate, and just kinda made what estimates we could until it fell out just right. Sometimes (often) it just takes trial and error.
              $endgroup$
              – AJY
              1 hour ago














            4












            4








            4





            $begingroup$

            Remember that in induction proofs, we start by assuming that the claim we're trying to prove is true for $n$, and then conclude that it must also be true for $n + 1$. In this case, we start with the induction assumption
            $$sum_{k = 1}^{n} frac{k}{k + 1} leq n - frac{1}{n + 1},$$
            and want to end with the conclusion that $$sum_{k = 1}^{n + 1} frac{k}{k + 1} leq n + 1 - frac{1}{n + 2} .$$
            Here's the argument written out in a bit more detail with commentary on each step.
            begin{align*}
            sum_{k = 1}^{n + 1} frac{k}{k + 1} & = frac{n + 1}{n + 2} + sum_{k = 1}^{n} frac{k}{k + 1} & textrm{ (just writing out the sum)} \
            & leq frac{n + 1}{n + 2} + n - frac{1}{n + 1} & textrm{ (applying the induction hypothesis)} \
            & = 1 - frac{1}{n + 2} + n - frac{1}{n + 1} & textrm{ (rewriting $frac{n+ 1}{n + 2}$ as $frac{n + 2 - 1}{n + 2} = 1 - frac{1}{n + 2}$)} \
            & = n + 1 - frac{1}{n + 1} - frac{1}{n + 2} & textrm{ (regrouping)} \
            & = n + 1 - frac{(n + 2) + (n + 1)}{(n + 2)(n + 1)} & textrm{ (combining fractions)} \
            & = n + 1 - frac{2(n + 2) - 1}{(n + 1)(n + 2)} & textrm{ (regrouping the numerator)} \
            & = n + 1 - frac{2(n + 2)}{(n + 1)(n + 2)} + frac{1}{(n + 1)(n + 2)} & textrm{ (breaking the fraction back apart)} \
            & = n + 1 - frac{2}{n + 1} + frac{1}{(n + 1)(n + 2)} & textrm{ (simplifying the fraction)} \
            & leq n + 1 - frac{2}{n + 2} + frac{1}{(n + 1)(n + 2)} & textrm{ (we slightly modified the second-to-last summand)} \
            & leq n + 1 - frac{2}{n + 2} + frac{1}{n + 2} & textrm{ (modifying the last summand)} \
            & = n + 1 - frac{1}{n + 2} & textrm{ (combining the fractions)} .
            end{align*}

            So in summary, we began with the assumption that the claim held for $n$, and through some arithmetic trickery concluded that it therefore held for $n + 1$.






            share|cite|improve this answer











            $endgroup$



            Remember that in induction proofs, we start by assuming that the claim we're trying to prove is true for $n$, and then conclude that it must also be true for $n + 1$. In this case, we start with the induction assumption
            $$sum_{k = 1}^{n} frac{k}{k + 1} leq n - frac{1}{n + 1},$$
            and want to end with the conclusion that $$sum_{k = 1}^{n + 1} frac{k}{k + 1} leq n + 1 - frac{1}{n + 2} .$$
            Here's the argument written out in a bit more detail with commentary on each step.
            begin{align*}
            sum_{k = 1}^{n + 1} frac{k}{k + 1} & = frac{n + 1}{n + 2} + sum_{k = 1}^{n} frac{k}{k + 1} & textrm{ (just writing out the sum)} \
            & leq frac{n + 1}{n + 2} + n - frac{1}{n + 1} & textrm{ (applying the induction hypothesis)} \
            & = 1 - frac{1}{n + 2} + n - frac{1}{n + 1} & textrm{ (rewriting $frac{n+ 1}{n + 2}$ as $frac{n + 2 - 1}{n + 2} = 1 - frac{1}{n + 2}$)} \
            & = n + 1 - frac{1}{n + 1} - frac{1}{n + 2} & textrm{ (regrouping)} \
            & = n + 1 - frac{(n + 2) + (n + 1)}{(n + 2)(n + 1)} & textrm{ (combining fractions)} \
            & = n + 1 - frac{2(n + 2) - 1}{(n + 1)(n + 2)} & textrm{ (regrouping the numerator)} \
            & = n + 1 - frac{2(n + 2)}{(n + 1)(n + 2)} + frac{1}{(n + 1)(n + 2)} & textrm{ (breaking the fraction back apart)} \
            & = n + 1 - frac{2}{n + 1} + frac{1}{(n + 1)(n + 2)} & textrm{ (simplifying the fraction)} \
            & leq n + 1 - frac{2}{n + 2} + frac{1}{(n + 1)(n + 2)} & textrm{ (we slightly modified the second-to-last summand)} \
            & leq n + 1 - frac{2}{n + 2} + frac{1}{n + 2} & textrm{ (modifying the last summand)} \
            & = n + 1 - frac{1}{n + 2} & textrm{ (combining the fractions)} .
            end{align*}

            So in summary, we began with the assumption that the claim held for $n$, and through some arithmetic trickery concluded that it therefore held for $n + 1$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago

























            answered 2 hours ago









            AJYAJY

            4,19521128




            4,19521128








            • 1




              $begingroup$
              This is an amazing break down, could explain your second-to-last and third-to-last steps where you modify the summand? Oh I think I might understand, is it to make the inequality <= ?
              $endgroup$
              – Brownie
              1 hour ago








            • 1




              $begingroup$
              @Brownie In the third-to-last step, we observe that $n + 1 leq n + 2$, so $frac{2}{n + 1} geq frac{2}{n + 2}$, so $- frac{2}{n + 1} leq - frac{2}{n + 2}$. For the second-to-last step, we can see that $frac{1}{n + 2} = frac{n + 1}{(n + 1)(n + 2)} = (n + 1) frac{1}{(n + 1)(n + 2)} geq frac{1}{(n + 1)(n + 2)}$.
              $endgroup$
              – AJY
              1 hour ago








            • 1




              $begingroup$
              So is this a step you saw you could implement to get the final equal to $ n + 1 - frac{1}{n + 2} $ ? Or is there something that would push you towards doing this?
              $endgroup$
              – Brownie
              1 hour ago








            • 2




              $begingroup$
              @Brownie The short answer is that these kinds of tricks come easier with practice. We wanted to get to a very particular estimate, and just kinda made what estimates we could until it fell out just right. Sometimes (often) it just takes trial and error.
              $endgroup$
              – AJY
              1 hour ago














            • 1




              $begingroup$
              This is an amazing break down, could explain your second-to-last and third-to-last steps where you modify the summand? Oh I think I might understand, is it to make the inequality <= ?
              $endgroup$
              – Brownie
              1 hour ago








            • 1




              $begingroup$
              @Brownie In the third-to-last step, we observe that $n + 1 leq n + 2$, so $frac{2}{n + 1} geq frac{2}{n + 2}$, so $- frac{2}{n + 1} leq - frac{2}{n + 2}$. For the second-to-last step, we can see that $frac{1}{n + 2} = frac{n + 1}{(n + 1)(n + 2)} = (n + 1) frac{1}{(n + 1)(n + 2)} geq frac{1}{(n + 1)(n + 2)}$.
              $endgroup$
              – AJY
              1 hour ago








            • 1




              $begingroup$
              So is this a step you saw you could implement to get the final equal to $ n + 1 - frac{1}{n + 2} $ ? Or is there something that would push you towards doing this?
              $endgroup$
              – Brownie
              1 hour ago








            • 2




              $begingroup$
              @Brownie The short answer is that these kinds of tricks come easier with practice. We wanted to get to a very particular estimate, and just kinda made what estimates we could until it fell out just right. Sometimes (often) it just takes trial and error.
              $endgroup$
              – AJY
              1 hour ago








            1




            1




            $begingroup$
            This is an amazing break down, could explain your second-to-last and third-to-last steps where you modify the summand? Oh I think I might understand, is it to make the inequality <= ?
            $endgroup$
            – Brownie
            1 hour ago






            $begingroup$
            This is an amazing break down, could explain your second-to-last and third-to-last steps where you modify the summand? Oh I think I might understand, is it to make the inequality <= ?
            $endgroup$
            – Brownie
            1 hour ago






            1




            1




            $begingroup$
            @Brownie In the third-to-last step, we observe that $n + 1 leq n + 2$, so $frac{2}{n + 1} geq frac{2}{n + 2}$, so $- frac{2}{n + 1} leq - frac{2}{n + 2}$. For the second-to-last step, we can see that $frac{1}{n + 2} = frac{n + 1}{(n + 1)(n + 2)} = (n + 1) frac{1}{(n + 1)(n + 2)} geq frac{1}{(n + 1)(n + 2)}$.
            $endgroup$
            – AJY
            1 hour ago






            $begingroup$
            @Brownie In the third-to-last step, we observe that $n + 1 leq n + 2$, so $frac{2}{n + 1} geq frac{2}{n + 2}$, so $- frac{2}{n + 1} leq - frac{2}{n + 2}$. For the second-to-last step, we can see that $frac{1}{n + 2} = frac{n + 1}{(n + 1)(n + 2)} = (n + 1) frac{1}{(n + 1)(n + 2)} geq frac{1}{(n + 1)(n + 2)}$.
            $endgroup$
            – AJY
            1 hour ago






            1




            1




            $begingroup$
            So is this a step you saw you could implement to get the final equal to $ n + 1 - frac{1}{n + 2} $ ? Or is there something that would push you towards doing this?
            $endgroup$
            – Brownie
            1 hour ago






            $begingroup$
            So is this a step you saw you could implement to get the final equal to $ n + 1 - frac{1}{n + 2} $ ? Or is there something that would push you towards doing this?
            $endgroup$
            – Brownie
            1 hour ago






            2




            2




            $begingroup$
            @Brownie The short answer is that these kinds of tricks come easier with practice. We wanted to get to a very particular estimate, and just kinda made what estimates we could until it fell out just right. Sometimes (often) it just takes trial and error.
            $endgroup$
            – AJY
            1 hour ago




            $begingroup$
            @Brownie The short answer is that these kinds of tricks come easier with practice. We wanted to get to a very particular estimate, and just kinda made what estimates we could until it fell out just right. Sometimes (often) it just takes trial and error.
            $endgroup$
            – AJY
            1 hour ago











            1












            $begingroup$

            Regarding the first inequality you're asking about, they are actually adding the term $frac{n+1}{n+2}$ to both sides, but on the left hand side it is added by increasing the upper limit in the sum by $1$, and to not change the inequality you have to add $frac{n+1}{n+2}$ to the right hand side.



            The subsequent steps in the proof you posted are just rearrangements of fractions using algebra, nothing actually changes there until the last line. Then they just use the fact that $frac{1}{(n+1)(n+2)} leq frac{1}{n+2}$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Cool just last quick question, for the left hand side if instead of increasing the upper limit of the sum by 1, I added $frac{n+1}{n+2}$ to both sides, that would be the same?
              $endgroup$
              – Brownie
              2 hours ago










            • $begingroup$
              Yes, that would be the same.
              $endgroup$
              – Thomas Fjærvik
              2 hours ago
















            1












            $begingroup$

            Regarding the first inequality you're asking about, they are actually adding the term $frac{n+1}{n+2}$ to both sides, but on the left hand side it is added by increasing the upper limit in the sum by $1$, and to not change the inequality you have to add $frac{n+1}{n+2}$ to the right hand side.



            The subsequent steps in the proof you posted are just rearrangements of fractions using algebra, nothing actually changes there until the last line. Then they just use the fact that $frac{1}{(n+1)(n+2)} leq frac{1}{n+2}$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Cool just last quick question, for the left hand side if instead of increasing the upper limit of the sum by 1, I added $frac{n+1}{n+2}$ to both sides, that would be the same?
              $endgroup$
              – Brownie
              2 hours ago










            • $begingroup$
              Yes, that would be the same.
              $endgroup$
              – Thomas Fjærvik
              2 hours ago














            1












            1








            1





            $begingroup$

            Regarding the first inequality you're asking about, they are actually adding the term $frac{n+1}{n+2}$ to both sides, but on the left hand side it is added by increasing the upper limit in the sum by $1$, and to not change the inequality you have to add $frac{n+1}{n+2}$ to the right hand side.



            The subsequent steps in the proof you posted are just rearrangements of fractions using algebra, nothing actually changes there until the last line. Then they just use the fact that $frac{1}{(n+1)(n+2)} leq frac{1}{n+2}$






            share|cite|improve this answer









            $endgroup$



            Regarding the first inequality you're asking about, they are actually adding the term $frac{n+1}{n+2}$ to both sides, but on the left hand side it is added by increasing the upper limit in the sum by $1$, and to not change the inequality you have to add $frac{n+1}{n+2}$ to the right hand side.



            The subsequent steps in the proof you posted are just rearrangements of fractions using algebra, nothing actually changes there until the last line. Then they just use the fact that $frac{1}{(n+1)(n+2)} leq frac{1}{n+2}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            Thomas FjærvikThomas Fjærvik

            2038




            2038












            • $begingroup$
              Cool just last quick question, for the left hand side if instead of increasing the upper limit of the sum by 1, I added $frac{n+1}{n+2}$ to both sides, that would be the same?
              $endgroup$
              – Brownie
              2 hours ago










            • $begingroup$
              Yes, that would be the same.
              $endgroup$
              – Thomas Fjærvik
              2 hours ago


















            • $begingroup$
              Cool just last quick question, for the left hand side if instead of increasing the upper limit of the sum by 1, I added $frac{n+1}{n+2}$ to both sides, that would be the same?
              $endgroup$
              – Brownie
              2 hours ago










            • $begingroup$
              Yes, that would be the same.
              $endgroup$
              – Thomas Fjærvik
              2 hours ago
















            $begingroup$
            Cool just last quick question, for the left hand side if instead of increasing the upper limit of the sum by 1, I added $frac{n+1}{n+2}$ to both sides, that would be the same?
            $endgroup$
            – Brownie
            2 hours ago




            $begingroup$
            Cool just last quick question, for the left hand side if instead of increasing the upper limit of the sum by 1, I added $frac{n+1}{n+2}$ to both sides, that would be the same?
            $endgroup$
            – Brownie
            2 hours ago












            $begingroup$
            Yes, that would be the same.
            $endgroup$
            – Thomas Fjærvik
            2 hours ago




            $begingroup$
            Yes, that would be the same.
            $endgroup$
            – Thomas Fjærvik
            2 hours ago











            1












            $begingroup$

            If $aleq b $, then $a+cleq b+c $ for any $cin Bbb R $. By assumption, we have $$sum_{k=1}^{n}frac k{k+1}leq n-frac 1{n+1}.$$ Now add $dfrac{n+1}{n+2}$ on both sides, i.e., $$sum_{k=1}^{n}frac k{k+1}+frac{n+1}{n+2}leq n-frac 1{n+1}+frac{n+1}{n+2}.$$ Note that $$sum_{k=1}^{n}frac k{k+1}+frac{n+1}{n+2}=sum_{k=1}^{n+1}frac k{k+1}.$$ Hence we have $$sum_{k=1}^{n+1}frac k{k+1}leq n-frac 1{n+1}+frac{n+1}{n+2}.$$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              If $aleq b $, then $a+cleq b+c $ for any $cin Bbb R $. By assumption, we have $$sum_{k=1}^{n}frac k{k+1}leq n-frac 1{n+1}.$$ Now add $dfrac{n+1}{n+2}$ on both sides, i.e., $$sum_{k=1}^{n}frac k{k+1}+frac{n+1}{n+2}leq n-frac 1{n+1}+frac{n+1}{n+2}.$$ Note that $$sum_{k=1}^{n}frac k{k+1}+frac{n+1}{n+2}=sum_{k=1}^{n+1}frac k{k+1}.$$ Hence we have $$sum_{k=1}^{n+1}frac k{k+1}leq n-frac 1{n+1}+frac{n+1}{n+2}.$$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                If $aleq b $, then $a+cleq b+c $ for any $cin Bbb R $. By assumption, we have $$sum_{k=1}^{n}frac k{k+1}leq n-frac 1{n+1}.$$ Now add $dfrac{n+1}{n+2}$ on both sides, i.e., $$sum_{k=1}^{n}frac k{k+1}+frac{n+1}{n+2}leq n-frac 1{n+1}+frac{n+1}{n+2}.$$ Note that $$sum_{k=1}^{n}frac k{k+1}+frac{n+1}{n+2}=sum_{k=1}^{n+1}frac k{k+1}.$$ Hence we have $$sum_{k=1}^{n+1}frac k{k+1}leq n-frac 1{n+1}+frac{n+1}{n+2}.$$






                share|cite|improve this answer











                $endgroup$



                If $aleq b $, then $a+cleq b+c $ for any $cin Bbb R $. By assumption, we have $$sum_{k=1}^{n}frac k{k+1}leq n-frac 1{n+1}.$$ Now add $dfrac{n+1}{n+2}$ on both sides, i.e., $$sum_{k=1}^{n}frac k{k+1}+frac{n+1}{n+2}leq n-frac 1{n+1}+frac{n+1}{n+2}.$$ Note that $$sum_{k=1}^{n}frac k{k+1}+frac{n+1}{n+2}=sum_{k=1}^{n+1}frac k{k+1}.$$ Hence we have $$sum_{k=1}^{n+1}frac k{k+1}leq n-frac 1{n+1}+frac{n+1}{n+2}.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 1 hour ago

























                answered 2 hours ago









                Thomas ShelbyThomas Shelby

                3,7192525




                3,7192525






























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