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I was able to find formulas for simpler expressions but I can't find the general formula for this one:

$\sum _{i=1}^{n} \frac {(-1)^i i}{(2i-1)(2i+1)}$

I don't see any particular trend that would help me in the first solutions.

$P(1) = -1/3, P(2) = -3/15, P(3) = -36/175 ...$

And if I expand the sum I also fail to come up with an answer. Any hints?

Thanks

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  • $\begingroup$ Did you really mean $21$ up there, or is it $2i$? $\endgroup$
    – barak manos
    Apr 25, 2016 at 4:11
  • $\begingroup$ nice catch, it was 2i. $\endgroup$
    – jrs
    Apr 25, 2016 at 4:12

2 Answers 2

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This is an example of a "telescoping" summation. Note that we can write

$$\begin{align} \sum_{i=1}^n\frac{(-1)^i i}{(2i-1)(2i+1)}&=\frac14 \sum_{i=1}^n \left(\frac{(-1)^i}{2i+1}-\frac{(-1)^{i-1}}{2i-1}\right)\\\\ &=\frac14 \sum_{i=1}^n \left(\frac{(-1)^i}{2i+1}-\frac{(-1)^{i-1}}{2(i-1)+1}\right)\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac14\left(\frac{(-1)^n}{2n+1}-1\right)} \end{align}$$

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$\frac{i}{(2i-1)(2i+1)}=\frac{1}{4}\left(\frac{1}{2i-1}+\frac{1}{2i+1}\right)$

Then you have

$-\frac{1}{4}\left(\frac{1}{2.1-1}+\frac{1}{2.1+1}\right)+\frac{1}{4}\left(\frac{1}{2.2-1}+\frac{1}{2.2+1}\right)+\ldots+$

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