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Find the sum of following series:

$\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...$

After some arrangement, I got below step:

$\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...=\sum\limits_{k=0}^{\infty}\frac{1}{2k+2}- \sum\limits_{k=0}^{\infty}\frac{1}{2k+3} $

Now, I have no idea how to find this difference of two series. Please help me. Thanks in advance.

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  • $\begingroup$ Expand $$\ln(1+x)$$ $\endgroup$ – lab bhattacharjee Dec 29 '15 at 15:45
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    $\begingroup$ The series on the left converges, and the two series on the right do not, so that approach may not be useful. Your series is $\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots$. Now have you seen something like this before? $\endgroup$ – André Nicolas Dec 29 '15 at 15:53
  • $\begingroup$ @lab battacharjee $ 1-\log 2$ $\endgroup$ – Siddhant Trivedi Dec 29 '15 at 15:55
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We have $$\sum_{k=1}^{\infty}\frac1{2k(2k+1)}=\sum_{k=1}^{\infty}\left(\frac1{2k}-\frac1{2k+1}\right)=\sum_{n=2}^{\infty}\frac{(-1)^n}{n}=1+\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=1-\ln2.$$ The last step follows from $\ln (1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}$ by setting $x=-1$.

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  • $\begingroup$ I got the answer before your post. Thanks $\endgroup$ – Siddhant Trivedi Dec 29 '15 at 16:03
  • $\begingroup$ How did you move from the telescoping expression to the alternating one? $\endgroup$ – zz20s Dec 29 '15 at 16:15
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    $\begingroup$ @zz20s The $k$th term in the first expression is the sum of $2k$th and $(2k+1)$th terms in the second (e.g. $\frac12-\frac13=\frac{(-1)^2}2 +\frac{(-1)^3}3$). $\endgroup$ – Start wearing purple Dec 29 '15 at 16:39
  • $\begingroup$ That makes sense, thank you. $\endgroup$ – zz20s Dec 29 '15 at 16:40
  • $\begingroup$ @zz20s I converted each terms appear in sum to factorial form. $\endgroup$ – Siddhant Trivedi Dec 30 '15 at 6:15
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The general term for the above series is 1/(2r)(2r+1). This can be broken into partial fractions. Write the numerator as 2r+1-2r and thus this general term can be written as 1/(2r+1) - 1/(2r). sum the terms from r=1 to n, and then take the answer in a limit n approaching infinity.

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  • $\begingroup$ Please use MathJax in your posts. $\endgroup$ – Kamil Jarosz Dec 29 '15 at 15:57

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