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I am struggling to find the sum of the following series:

$$\sum_{n=1} ^{\infty} \frac{(-1)^n}{(n+1)(n+3)(n+5)}.$$

It seems as though it should be a straightforward telescoping series. I attempted to solve it in the usual way (via partial fractions), but the alternating sign makes the sum so that one cannot cancel out fractions to result in a finite sum of fractions. I know that the series converges by the alternating sign test, and I check on WolframAlpha that the infinite sum converges to $-7/480$. Any thoughts on how to proceed?

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  • $\begingroup$ Once you have the partial fraction decomposition, have you tried to compute it directly as a function of the sum $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ (whose value is "known" to be $-\ln 2$ is you need it)? $\endgroup$
    – Clement C.
    Jun 23, 2015 at 19:28

3 Answers 3

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We have (through the residue theorem or by linear algebra) : $$ \frac{1}{(n+1)(n+3)(n+5)}=\frac{1}{8}\left(\frac{1}{n+1}-\frac{2}{n+3}+\frac{1}{n+5}\right) $$ as well as (by shifting the summation index): $$ \sum_{n\geq 1}\frac{(-1)^n}{n+1}=-1+\log 2, $$ $$ \sum_{n\geq 1}\frac{(-1)^n}{n+3}=-\frac{5}{6}+\log 2, $$ $$ \sum_{n\geq 1}\frac{(-1)^n}{n+5}=-\frac{47}{60}+\log 2. $$ Just combine them. $\log 2$ cancels out since $1-2+1=0$ (we have a meromorphic function that is $O\left(\frac{1}{|z|^2}\right)$ as $|z|\to +\infty$, hence the sum of its residues is necessary zero).

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  • $\begingroup$ Could you explain the last part in further detail please? Specifically, from "just combine..." and on. I've never used a meromorphic function before, so I'm fairly confused. Thank you! $\endgroup$
    – kathystehl
    Jun 23, 2015 at 21:04
  • $\begingroup$ @kathystehl: you may just ignore the part about the meromorphic function. It is just the intrinsic reason for which the $\log 2$ part has to cancel out, but you can check it in a straightforward way, i.e. $1-2+1=0$. $\endgroup$
    – Jack D'Aurizio
    Jun 23, 2015 at 21:06
  • $\begingroup$ Could you identify what you are referring to when you parallel the $log2$ cancelling out to the identity $1-2+1=0$? $\endgroup$
    – kathystehl
    Jun 23, 2015 at 21:35
  • $\begingroup$ About $(2)$, I am just saying that if you take the first line, minus twice the second line, plus the third line, there is no $\log$ in the final outcome. $\endgroup$
    – Jack D'Aurizio
    Jun 23, 2015 at 21:43
  • $\begingroup$ I see, my apologies for the confusion. One last question (I promise): how do we deal with the $1/8$ when we separate the series into three parts as you did? $\endgroup$
    – kathystehl
    Jun 23, 2015 at 21:47
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Rewrite the alternating sum as a difference of two infinite sums. $$\sum_{n=1} ^{\infty} \frac{(-1)^n}{(n+1)(n+3)(n+5)} = \sum_{n=1} ^{\infty} \frac{1}{(2n+1)(2n+3)(2n+5)}-\sum_{n=1} ^{\infty} \frac{1}{2n(2n+2)(2n+4)}$$ You'll probably wnat to convince yourself of the equality. Now you can use partial fraction decomposition to decompose $\frac{1}{(2n+1)(2n+3)(2n+5)}$ and $\frac{1}{2n(2n+2)(2n+4)}$ into three separate fractions each, giving you six infinite sums in total. From there you should be able to make some judicious cancellations and get your result. You may find one of my previous questions helpful at this point. Exact value of $\sum_{n=1}^\infty \frac{1}{n(n+k)(n+l)}$ for $k \in \Bbb{N}-\{0\}$ and $l \in \Bbb{N}-\{0,k\}$

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Partial Decomposition provides:

$S = \sum_{1}^{\infty} (-1)^n[\frac{1}{8(n+1)} - \frac{1}{4(n+3)}+\frac{1}{8(n+5)}]$

Splitting S_Even and S_Odd

$S_{Even} = \frac{1}{8.3} - \frac{1}{4.5}+\frac{1}{8.5}$

$S_{Odd} = -\frac{1}{8.2} +\frac{1}{4.4} - \frac{1}{8.4}$

Everything else cancels out

When you sum these you get $S = \dfrac{-7}{480}$

Provided I have not made any calculation error

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