Uncomfortable Series Calculations (not geometric nor telescoping): $sumlimits_{n=1}^{ infty }...












1












$begingroup$


I am trying to find the sum of the following infinite series:
$sum_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}$.
I tried to break it apart and solve like a telescoping series, but to no avail. Unless I have missed something major, it is definitely not a geometric series. The only way that I found the sum was to use Wolfram Alpha, which gave me the answer of 1. By just calculating and plugging in a bunch of numbers, I was able to see the answer tending towards 1, but I would like to know if there is an explicit way to calculate this. Any ideas?






sequences-and-series taylor-expansion







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  • 3




    $begingroup$
    Off the top of my head and without giving the problem any thought at all, a partial fraction decomposition seems promising.
    $endgroup$
    – Xander Henderson
    Jan 21 at 2:44






  • 1




    $begingroup$
    I checked and yes it's definitely a telescoping series. Just do the partial fraction decomposition.
    $endgroup$
    – Ben W
    Jan 21 at 2:49










  • $begingroup$
    $$(-1)^{n+1}frac{2n+1}{n(n+1)}=(-1)^{n+1}frac{n+(n+1)}{n(n+1)}=(-1)^{n+1}frac1{n+1}-(-1)^nfrac1n$$
    $endgroup$
    – Mark Viola
    Jan 21 at 5:08












  • $begingroup$
    See also: Does $frac{3}{1cdot 2} - frac{5}{2cdot 3} + frac{7}{3cdot 4} - ...$ Converges? and Sum of two harmonic alternating series.
    $endgroup$
    – Martin Sleziak
    Jan 21 at 15:23
















1












$begingroup$


I am trying to find the sum of the following infinite series:
$sum_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}$.
I tried to break it apart and solve like a telescoping series, but to no avail. Unless I have missed something major, it is definitely not a geometric series. The only way that I found the sum was to use Wolfram Alpha, which gave me the answer of 1. By just calculating and plugging in a bunch of numbers, I was able to see the answer tending towards 1, but I would like to know if there is an explicit way to calculate this. Any ideas?






sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Off the top of my head and without giving the problem any thought at all, a partial fraction decomposition seems promising.
    $endgroup$
    – Xander Henderson
    Jan 21 at 2:44






  • 1




    $begingroup$
    I checked and yes it's definitely a telescoping series. Just do the partial fraction decomposition.
    $endgroup$
    – Ben W
    Jan 21 at 2:49










  • $begingroup$
    $$(-1)^{n+1}frac{2n+1}{n(n+1)}=(-1)^{n+1}frac{n+(n+1)}{n(n+1)}=(-1)^{n+1}frac1{n+1}-(-1)^nfrac1n$$
    $endgroup$
    – Mark Viola
    Jan 21 at 5:08












  • $begingroup$
    See also: Does $frac{3}{1cdot 2} - frac{5}{2cdot 3} + frac{7}{3cdot 4} - ...$ Converges? and Sum of two harmonic alternating series.
    $endgroup$
    – Martin Sleziak
    Jan 21 at 15:23














1












1








1





$begingroup$


I am trying to find the sum of the following infinite series:
$sum_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}$.
I tried to break it apart and solve like a telescoping series, but to no avail. Unless I have missed something major, it is definitely not a geometric series. The only way that I found the sum was to use Wolfram Alpha, which gave me the answer of 1. By just calculating and plugging in a bunch of numbers, I was able to see the answer tending towards 1, but I would like to know if there is an explicit way to calculate this. Any ideas?






sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$




I am trying to find the sum of the following infinite series:
$sum_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}$.
I tried to break it apart and solve like a telescoping series, but to no avail. Unless I have missed something major, it is definitely not a geometric series. The only way that I found the sum was to use Wolfram Alpha, which gave me the answer of 1. By just calculating and plugging in a bunch of numbers, I was able to see the answer tending towards 1, but I would like to know if there is an explicit way to calculate this. Any ideas?






sequences-and-series taylor-expansion




sequences-and-series taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 15:23









Martin Sleziak

44.8k10119272




44.8k10119272










asked Jan 21 at 2:41









DevDragonDevDragon

83




83








  • 3




    $begingroup$
    Off the top of my head and without giving the problem any thought at all, a partial fraction decomposition seems promising.
    $endgroup$
    – Xander Henderson
    Jan 21 at 2:44






  • 1




    $begingroup$
    I checked and yes it's definitely a telescoping series. Just do the partial fraction decomposition.
    $endgroup$
    – Ben W
    Jan 21 at 2:49










  • $begingroup$
    $$(-1)^{n+1}frac{2n+1}{n(n+1)}=(-1)^{n+1}frac{n+(n+1)}{n(n+1)}=(-1)^{n+1}frac1{n+1}-(-1)^nfrac1n$$
    $endgroup$
    – Mark Viola
    Jan 21 at 5:08












  • $begingroup$
    See also: Does $frac{3}{1cdot 2} - frac{5}{2cdot 3} + frac{7}{3cdot 4} - ...$ Converges? and Sum of two harmonic alternating series.
    $endgroup$
    – Martin Sleziak
    Jan 21 at 15:23














  • 3




    $begingroup$
    Off the top of my head and without giving the problem any thought at all, a partial fraction decomposition seems promising.
    $endgroup$
    – Xander Henderson
    Jan 21 at 2:44






  • 1




    $begingroup$
    I checked and yes it's definitely a telescoping series. Just do the partial fraction decomposition.
    $endgroup$
    – Ben W
    Jan 21 at 2:49










  • $begingroup$
    $$(-1)^{n+1}frac{2n+1}{n(n+1)}=(-1)^{n+1}frac{n+(n+1)}{n(n+1)}=(-1)^{n+1}frac1{n+1}-(-1)^nfrac1n$$
    $endgroup$
    – Mark Viola
    Jan 21 at 5:08












  • $begingroup$
    See also: Does $frac{3}{1cdot 2} - frac{5}{2cdot 3} + frac{7}{3cdot 4} - ...$ Converges? and Sum of two harmonic alternating series.
    $endgroup$
    – Martin Sleziak
    Jan 21 at 15:23








3




3




$begingroup$
Off the top of my head and without giving the problem any thought at all, a partial fraction decomposition seems promising.
$endgroup$
– Xander Henderson
Jan 21 at 2:44




$begingroup$
Off the top of my head and without giving the problem any thought at all, a partial fraction decomposition seems promising.
$endgroup$
– Xander Henderson
Jan 21 at 2:44




1




1




$begingroup$
I checked and yes it's definitely a telescoping series. Just do the partial fraction decomposition.
$endgroup$
– Ben W
Jan 21 at 2:49




$begingroup$
I checked and yes it's definitely a telescoping series. Just do the partial fraction decomposition.
$endgroup$
– Ben W
Jan 21 at 2:49












$begingroup$
$$(-1)^{n+1}frac{2n+1}{n(n+1)}=(-1)^{n+1}frac{n+(n+1)}{n(n+1)}=(-1)^{n+1}frac1{n+1}-(-1)^nfrac1n$$
$endgroup$
– Mark Viola
Jan 21 at 5:08






$begingroup$
$$(-1)^{n+1}frac{2n+1}{n(n+1)}=(-1)^{n+1}frac{n+(n+1)}{n(n+1)}=(-1)^{n+1}frac1{n+1}-(-1)^nfrac1n$$
$endgroup$
– Mark Viola
Jan 21 at 5:08














$begingroup$
See also: Does $frac{3}{1cdot 2} - frac{5}{2cdot 3} + frac{7}{3cdot 4} - ...$ Converges? and Sum of two harmonic alternating series.
$endgroup$
– Martin Sleziak
Jan 21 at 15:23




$begingroup$
See also: Does $frac{3}{1cdot 2} - frac{5}{2cdot 3} + frac{7}{3cdot 4} - ...$ Converges? and Sum of two harmonic alternating series.
$endgroup$
– Martin Sleziak
Jan 21 at 15:23










1 Answer
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$begingroup$

Let's break the sum into two parts:



$sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}=sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n}{n(n+1)}+sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{1}{n(n+1)} tag1$



The first sum is equal to:



$-2sumlimits_{n=1}^{ infty } frac{(-1)^{n}}{n+1}=big(-2sumlimits_{n=0}^{ infty } frac{(-1)^{n}}{n+1}x^{n+1}+2big)_{x=1}=-2ln2+2 tag2$



At the second one using partial fraction decomposition:



$sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{1}{n(n+1)}= sumlimits_{n=1}^{ infty } frac{(-1)^{n+1}}{n}- sumlimits_{n=1}^{ infty } frac{(-1)^{n+1}}{n+1}tag3$



Start both sums from zero and using the same fact (Taylor series of ln(x+1)):



$sumlimits_{n=0}^{ infty } frac{(-1)^{n+2}}{n+1}- sumlimits_{n=0}^{ infty } frac{(-1)^{n+1}}{n+1}-1=ln2+ln2 -1tag4$



Summarize the results of (2) and (4) we get:



$sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}=1tag5$






share|cite|improve this answer











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    1 Answer
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    active

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    active

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    0












    $begingroup$

    Let's break the sum into two parts:



    $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}=sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n}{n(n+1)}+sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{1}{n(n+1)} tag1$



    The first sum is equal to:



    $-2sumlimits_{n=1}^{ infty } frac{(-1)^{n}}{n+1}=big(-2sumlimits_{n=0}^{ infty } frac{(-1)^{n}}{n+1}x^{n+1}+2big)_{x=1}=-2ln2+2 tag2$



    At the second one using partial fraction decomposition:



    $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{1}{n(n+1)}= sumlimits_{n=1}^{ infty } frac{(-1)^{n+1}}{n}- sumlimits_{n=1}^{ infty } frac{(-1)^{n+1}}{n+1}tag3$



    Start both sums from zero and using the same fact (Taylor series of ln(x+1)):



    $sumlimits_{n=0}^{ infty } frac{(-1)^{n+2}}{n+1}- sumlimits_{n=0}^{ infty } frac{(-1)^{n+1}}{n+1}-1=ln2+ln2 -1tag4$



    Summarize the results of (2) and (4) we get:



    $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}=1tag5$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let's break the sum into two parts:



      $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}=sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n}{n(n+1)}+sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{1}{n(n+1)} tag1$



      The first sum is equal to:



      $-2sumlimits_{n=1}^{ infty } frac{(-1)^{n}}{n+1}=big(-2sumlimits_{n=0}^{ infty } frac{(-1)^{n}}{n+1}x^{n+1}+2big)_{x=1}=-2ln2+2 tag2$



      At the second one using partial fraction decomposition:



      $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{1}{n(n+1)}= sumlimits_{n=1}^{ infty } frac{(-1)^{n+1}}{n}- sumlimits_{n=1}^{ infty } frac{(-1)^{n+1}}{n+1}tag3$



      Start both sums from zero and using the same fact (Taylor series of ln(x+1)):



      $sumlimits_{n=0}^{ infty } frac{(-1)^{n+2}}{n+1}- sumlimits_{n=0}^{ infty } frac{(-1)^{n+1}}{n+1}-1=ln2+ln2 -1tag4$



      Summarize the results of (2) and (4) we get:



      $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}=1tag5$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let's break the sum into two parts:



        $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}=sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n}{n(n+1)}+sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{1}{n(n+1)} tag1$



        The first sum is equal to:



        $-2sumlimits_{n=1}^{ infty } frac{(-1)^{n}}{n+1}=big(-2sumlimits_{n=0}^{ infty } frac{(-1)^{n}}{n+1}x^{n+1}+2big)_{x=1}=-2ln2+2 tag2$



        At the second one using partial fraction decomposition:



        $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{1}{n(n+1)}= sumlimits_{n=1}^{ infty } frac{(-1)^{n+1}}{n}- sumlimits_{n=1}^{ infty } frac{(-1)^{n+1}}{n+1}tag3$



        Start both sums from zero and using the same fact (Taylor series of ln(x+1)):



        $sumlimits_{n=0}^{ infty } frac{(-1)^{n+2}}{n+1}- sumlimits_{n=0}^{ infty } frac{(-1)^{n+1}}{n+1}-1=ln2+ln2 -1tag4$



        Summarize the results of (2) and (4) we get:



        $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}=1tag5$






        share|cite|improve this answer











        $endgroup$



        Let's break the sum into two parts:



        $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}=sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n}{n(n+1)}+sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{1}{n(n+1)} tag1$



        The first sum is equal to:



        $-2sumlimits_{n=1}^{ infty } frac{(-1)^{n}}{n+1}=big(-2sumlimits_{n=0}^{ infty } frac{(-1)^{n}}{n+1}x^{n+1}+2big)_{x=1}=-2ln2+2 tag2$



        At the second one using partial fraction decomposition:



        $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{1}{n(n+1)}= sumlimits_{n=1}^{ infty } frac{(-1)^{n+1}}{n}- sumlimits_{n=1}^{ infty } frac{(-1)^{n+1}}{n+1}tag3$



        Start both sums from zero and using the same fact (Taylor series of ln(x+1)):



        $sumlimits_{n=0}^{ infty } frac{(-1)^{n+2}}{n+1}- sumlimits_{n=0}^{ infty } frac{(-1)^{n+1}}{n+1}-1=ln2+ln2 -1tag4$



        Summarize the results of (2) and (4) we get:



        $sumlimits_{n=1}^{ infty } (-1)^{n+1}frac{2n+1}{n(n+1)}=1tag5$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 21 at 4:56

























        answered Jan 21 at 4:03









        JV.StalkerJV.Stalker

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