Summation of series and Taylor series are giving different results












1












$begingroup$


$A = frac{1}{2} + frac{1}{4} + frac{1}{6} + cdots$
$B = 1 + frac{1}{3} + frac{1}{5} + frac{1}{7} + cdots$
$A + B = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + cdots$
$2A = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + cdots$
$2A = A + B$
$A - B = 0$
$0 = 1 - frac{1}{2} + frac{1}{3} - frac{1}{4} + frac{1}{5} - frac{1}{6} + cdots$



Using Taylor expansion for $ln(1+x)$
$ln(2) = 1 -frac{1}{2} + frac{1}{3} - frac{1}{4} + frac{1}{5} - frac{1}{6} + cdots$



Where did I go wrong ?






sequences-and-series summation taylor-expansion







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You are manipulating divergent series, hence the confusion. Take a look on Riemann theorem.
    $endgroup$
    – rafa11111
    Dec 19 '18 at 17:33






  • 1




    $begingroup$
    Look here for a detailed explanation on why this can't be done.
    $endgroup$
    – Viktor Glombik
    Dec 19 '18 at 17:36












  • $begingroup$
    @ViktorGlombik This Mathologer video is also suitable for this question!
    $endgroup$
    – rafa11111
    Dec 19 '18 at 17:39










  • $begingroup$
    You're ignoring the fact that $A$ and $B$ are not finite numbers. Essentially, they are both $infty$ so $A-B$ is meaningless and cancellation cannot be performed as you have done.
    $endgroup$
    – MPW
    Dec 19 '18 at 17:46






  • 2




    $begingroup$
    Look at the partial sums $A_N=sum_{n=1}^{N} frac{1}{2n}$ and $B_N=sum_{n=1}^N frac{1}{2n-1}$. While $A_N+B_N =sum_{n=1}^{2N} frac1n$, we have $A_N-B_N=sum_{n=N+1}^{2N}frac1n ne 0$. In fact, we have $lim_{Ntoinfty}(A_N-B_N)=log(2)$
    $endgroup$
    – Mark Viola
    Dec 19 '18 at 18:20


















1












$begingroup$


$A = frac{1}{2} + frac{1}{4} + frac{1}{6} + cdots$
$B = 1 + frac{1}{3} + frac{1}{5} + frac{1}{7} + cdots$
$A + B = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + cdots$
$2A = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + cdots$
$2A = A + B$
$A - B = 0$
$0 = 1 - frac{1}{2} + frac{1}{3} - frac{1}{4} + frac{1}{5} - frac{1}{6} + cdots$



Using Taylor expansion for $ln(1+x)$
$ln(2) = 1 -frac{1}{2} + frac{1}{3} - frac{1}{4} + frac{1}{5} - frac{1}{6} + cdots$



Where did I go wrong ?






sequences-and-series summation taylor-expansion







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You are manipulating divergent series, hence the confusion. Take a look on Riemann theorem.
    $endgroup$
    – rafa11111
    Dec 19 '18 at 17:33






  • 1




    $begingroup$
    Look here for a detailed explanation on why this can't be done.
    $endgroup$
    – Viktor Glombik
    Dec 19 '18 at 17:36












  • $begingroup$
    @ViktorGlombik This Mathologer video is also suitable for this question!
    $endgroup$
    – rafa11111
    Dec 19 '18 at 17:39










  • $begingroup$
    You're ignoring the fact that $A$ and $B$ are not finite numbers. Essentially, they are both $infty$ so $A-B$ is meaningless and cancellation cannot be performed as you have done.
    $endgroup$
    – MPW
    Dec 19 '18 at 17:46






  • 2




    $begingroup$
    Look at the partial sums $A_N=sum_{n=1}^{N} frac{1}{2n}$ and $B_N=sum_{n=1}^N frac{1}{2n-1}$. While $A_N+B_N =sum_{n=1}^{2N} frac1n$, we have $A_N-B_N=sum_{n=N+1}^{2N}frac1n ne 0$. In fact, we have $lim_{Ntoinfty}(A_N-B_N)=log(2)$
    $endgroup$
    – Mark Viola
    Dec 19 '18 at 18:20
















1












1








1





$begingroup$


$A = frac{1}{2} + frac{1}{4} + frac{1}{6} + cdots$
$B = 1 + frac{1}{3} + frac{1}{5} + frac{1}{7} + cdots$
$A + B = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + cdots$
$2A = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + cdots$
$2A = A + B$
$A - B = 0$
$0 = 1 - frac{1}{2} + frac{1}{3} - frac{1}{4} + frac{1}{5} - frac{1}{6} + cdots$



Using Taylor expansion for $ln(1+x)$
$ln(2) = 1 -frac{1}{2} + frac{1}{3} - frac{1}{4} + frac{1}{5} - frac{1}{6} + cdots$



Where did I go wrong ?






sequences-and-series summation taylor-expansion







share|cite|improve this question











$endgroup$




$A = frac{1}{2} + frac{1}{4} + frac{1}{6} + cdots$
$B = 1 + frac{1}{3} + frac{1}{5} + frac{1}{7} + cdots$
$A + B = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + cdots$
$2A = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + cdots$
$2A = A + B$
$A - B = 0$
$0 = 1 - frac{1}{2} + frac{1}{3} - frac{1}{4} + frac{1}{5} - frac{1}{6} + cdots$



Using Taylor expansion for $ln(1+x)$
$ln(2) = 1 -frac{1}{2} + frac{1}{3} - frac{1}{4} + frac{1}{5} - frac{1}{6} + cdots$



Where did I go wrong ?






sequences-and-series summation taylor-expansion




sequences-and-series summation taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 18:23









Davide Giraudo

126k16150261




126k16150261










asked Dec 19 '18 at 17:30









Akshat BaranwalAkshat Baranwal

84




84








  • 2




    $begingroup$
    You are manipulating divergent series, hence the confusion. Take a look on Riemann theorem.
    $endgroup$
    – rafa11111
    Dec 19 '18 at 17:33






  • 1




    $begingroup$
    Look here for a detailed explanation on why this can't be done.
    $endgroup$
    – Viktor Glombik
    Dec 19 '18 at 17:36












  • $begingroup$
    @ViktorGlombik This Mathologer video is also suitable for this question!
    $endgroup$
    – rafa11111
    Dec 19 '18 at 17:39










  • $begingroup$
    You're ignoring the fact that $A$ and $B$ are not finite numbers. Essentially, they are both $infty$ so $A-B$ is meaningless and cancellation cannot be performed as you have done.
    $endgroup$
    – MPW
    Dec 19 '18 at 17:46






  • 2




    $begingroup$
    Look at the partial sums $A_N=sum_{n=1}^{N} frac{1}{2n}$ and $B_N=sum_{n=1}^N frac{1}{2n-1}$. While $A_N+B_N =sum_{n=1}^{2N} frac1n$, we have $A_N-B_N=sum_{n=N+1}^{2N}frac1n ne 0$. In fact, we have $lim_{Ntoinfty}(A_N-B_N)=log(2)$
    $endgroup$
    – Mark Viola
    Dec 19 '18 at 18:20
















  • 2




    $begingroup$
    You are manipulating divergent series, hence the confusion. Take a look on Riemann theorem.
    $endgroup$
    – rafa11111
    Dec 19 '18 at 17:33






  • 1




    $begingroup$
    Look here for a detailed explanation on why this can't be done.
    $endgroup$
    – Viktor Glombik
    Dec 19 '18 at 17:36












  • $begingroup$
    @ViktorGlombik This Mathologer video is also suitable for this question!
    $endgroup$
    – rafa11111
    Dec 19 '18 at 17:39










  • $begingroup$
    You're ignoring the fact that $A$ and $B$ are not finite numbers. Essentially, they are both $infty$ so $A-B$ is meaningless and cancellation cannot be performed as you have done.
    $endgroup$
    – MPW
    Dec 19 '18 at 17:46






  • 2




    $begingroup$
    Look at the partial sums $A_N=sum_{n=1}^{N} frac{1}{2n}$ and $B_N=sum_{n=1}^N frac{1}{2n-1}$. While $A_N+B_N =sum_{n=1}^{2N} frac1n$, we have $A_N-B_N=sum_{n=N+1}^{2N}frac1n ne 0$. In fact, we have $lim_{Ntoinfty}(A_N-B_N)=log(2)$
    $endgroup$
    – Mark Viola
    Dec 19 '18 at 18:20










2




2




$begingroup$
You are manipulating divergent series, hence the confusion. Take a look on Riemann theorem.
$endgroup$
– rafa11111
Dec 19 '18 at 17:33




$begingroup$
You are manipulating divergent series, hence the confusion. Take a look on Riemann theorem.
$endgroup$
– rafa11111
Dec 19 '18 at 17:33




1




1




$begingroup$
Look here for a detailed explanation on why this can't be done.
$endgroup$
– Viktor Glombik
Dec 19 '18 at 17:36






$begingroup$
Look here for a detailed explanation on why this can't be done.
$endgroup$
– Viktor Glombik
Dec 19 '18 at 17:36














$begingroup$
@ViktorGlombik This Mathologer video is also suitable for this question!
$endgroup$
– rafa11111
Dec 19 '18 at 17:39




$begingroup$
@ViktorGlombik This Mathologer video is also suitable for this question!
$endgroup$
– rafa11111
Dec 19 '18 at 17:39












$begingroup$
You're ignoring the fact that $A$ and $B$ are not finite numbers. Essentially, they are both $infty$ so $A-B$ is meaningless and cancellation cannot be performed as you have done.
$endgroup$
– MPW
Dec 19 '18 at 17:46




$begingroup$
You're ignoring the fact that $A$ and $B$ are not finite numbers. Essentially, they are both $infty$ so $A-B$ is meaningless and cancellation cannot be performed as you have done.
$endgroup$
– MPW
Dec 19 '18 at 17:46




2




2




$begingroup$
Look at the partial sums $A_N=sum_{n=1}^{N} frac{1}{2n}$ and $B_N=sum_{n=1}^N frac{1}{2n-1}$. While $A_N+B_N =sum_{n=1}^{2N} frac1n$, we have $A_N-B_N=sum_{n=N+1}^{2N}frac1n ne 0$. In fact, we have $lim_{Ntoinfty}(A_N-B_N)=log(2)$
$endgroup$
– Mark Viola
Dec 19 '18 at 18:20






$begingroup$
Look at the partial sums $A_N=sum_{n=1}^{N} frac{1}{2n}$ and $B_N=sum_{n=1}^N frac{1}{2n-1}$. While $A_N+B_N =sum_{n=1}^{2N} frac1n$, we have $A_N-B_N=sum_{n=N+1}^{2N}frac1n ne 0$. In fact, we have $lim_{Ntoinfty}(A_N-B_N)=log(2)$
$endgroup$
– Mark Viola
Dec 19 '18 at 18:20












2 Answers
2






active

oldest

votes


















1












$begingroup$

What you did wrong was to assume that you can deal with divergent series as if they were convergent. If you say that



$$
A=frac12+frac14+frac16+cdots
$$



then $A$ is a name of the series, but it is not a number (since the series diverges).



Therefore
$$
1+frac12+frac13+frac14+cdots
$$
is just another (divergent) series and it is an unfortunate option to call it $2A$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    As others have already pointed out, the series $sum_{n=1}^infty frac1n$ and $sum_{n=1}^infty frac1{2n-1}$ diverge and manipulating divergent series is not legitimate.




    Another way to analyze this is to look at the partial sums $A_N=sum_{n=1}^{N} frac{1}{2n}$ and $B_N=sum_{n=1}^N frac{1}{2n-1}$.



    While $A_N+B_N =sum_{n=1}^{2N} frac1n$, we have $A_N-B_N=-sum_{n=N+1}^{2N}frac1n ne 0$ (In fact, we have $lim_{Ntoinfty}(A_N-B_N)=log(1/2)$).






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      What you did wrong was to assume that you can deal with divergent series as if they were convergent. If you say that



      $$
      A=frac12+frac14+frac16+cdots
      $$



      then $A$ is a name of the series, but it is not a number (since the series diverges).



      Therefore
      $$
      1+frac12+frac13+frac14+cdots
      $$
      is just another (divergent) series and it is an unfortunate option to call it $2A$.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        What you did wrong was to assume that you can deal with divergent series as if they were convergent. If you say that



        $$
        A=frac12+frac14+frac16+cdots
        $$



        then $A$ is a name of the series, but it is not a number (since the series diverges).



        Therefore
        $$
        1+frac12+frac13+frac14+cdots
        $$
        is just another (divergent) series and it is an unfortunate option to call it $2A$.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          What you did wrong was to assume that you can deal with divergent series as if they were convergent. If you say that



          $$
          A=frac12+frac14+frac16+cdots
          $$



          then $A$ is a name of the series, but it is not a number (since the series diverges).



          Therefore
          $$
          1+frac12+frac13+frac14+cdots
          $$
          is just another (divergent) series and it is an unfortunate option to call it $2A$.






          share|cite|improve this answer











          $endgroup$



          What you did wrong was to assume that you can deal with divergent series as if they were convergent. If you say that



          $$
          A=frac12+frac14+frac16+cdots
          $$



          then $A$ is a name of the series, but it is not a number (since the series diverges).



          Therefore
          $$
          1+frac12+frac13+frac14+cdots
          $$
          is just another (divergent) series and it is an unfortunate option to call it $2A$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 17:38









          amWhy

          1




          1










          answered Dec 19 '18 at 17:34









          José Carlos SantosJosé Carlos Santos

          157k22126227




          157k22126227























              1












              $begingroup$

              As others have already pointed out, the series $sum_{n=1}^infty frac1n$ and $sum_{n=1}^infty frac1{2n-1}$ diverge and manipulating divergent series is not legitimate.




              Another way to analyze this is to look at the partial sums $A_N=sum_{n=1}^{N} frac{1}{2n}$ and $B_N=sum_{n=1}^N frac{1}{2n-1}$.



              While $A_N+B_N =sum_{n=1}^{2N} frac1n$, we have $A_N-B_N=-sum_{n=N+1}^{2N}frac1n ne 0$ (In fact, we have $lim_{Ntoinfty}(A_N-B_N)=log(1/2)$).






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                As others have already pointed out, the series $sum_{n=1}^infty frac1n$ and $sum_{n=1}^infty frac1{2n-1}$ diverge and manipulating divergent series is not legitimate.




                Another way to analyze this is to look at the partial sums $A_N=sum_{n=1}^{N} frac{1}{2n}$ and $B_N=sum_{n=1}^N frac{1}{2n-1}$.



                While $A_N+B_N =sum_{n=1}^{2N} frac1n$, we have $A_N-B_N=-sum_{n=N+1}^{2N}frac1n ne 0$ (In fact, we have $lim_{Ntoinfty}(A_N-B_N)=log(1/2)$).






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  As others have already pointed out, the series $sum_{n=1}^infty frac1n$ and $sum_{n=1}^infty frac1{2n-1}$ diverge and manipulating divergent series is not legitimate.




                  Another way to analyze this is to look at the partial sums $A_N=sum_{n=1}^{N} frac{1}{2n}$ and $B_N=sum_{n=1}^N frac{1}{2n-1}$.



                  While $A_N+B_N =sum_{n=1}^{2N} frac1n$, we have $A_N-B_N=-sum_{n=N+1}^{2N}frac1n ne 0$ (In fact, we have $lim_{Ntoinfty}(A_N-B_N)=log(1/2)$).






                  share|cite|improve this answer











                  $endgroup$



                  As others have already pointed out, the series $sum_{n=1}^infty frac1n$ and $sum_{n=1}^infty frac1{2n-1}$ diverge and manipulating divergent series is not legitimate.




                  Another way to analyze this is to look at the partial sums $A_N=sum_{n=1}^{N} frac{1}{2n}$ and $B_N=sum_{n=1}^N frac{1}{2n-1}$.



                  While $A_N+B_N =sum_{n=1}^{2N} frac1n$, we have $A_N-B_N=-sum_{n=N+1}^{2N}frac1n ne 0$ (In fact, we have $lim_{Ntoinfty}(A_N-B_N)=log(1/2)$).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 20 '18 at 1:34

























                  answered Dec 19 '18 at 18:27









                  Mark ViolaMark Viola

                  131k1275171




                  131k1275171






























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                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
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