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?


1 Answer 1


Let's break the sum into two parts:

$\sum\limits_{n=1}^{ \infty } (-1)^{n+1}\frac{2n+1}{n(n+1)}=\sum\limits_{n=1}^{ \infty } (-1)^{n+1}\frac{2n}{n(n+1)}+\sum\limits_{n=1}^{ \infty } (-1)^{n+1}\frac{1}{n(n+1)} \tag1$

The first sum is equal to:

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

At the second one using partial fraction decomposition:

$\sum\limits_{n=1}^{ \infty } (-1)^{n+1}\frac{1}{n(n+1)}= \sum\limits_{n=1}^{ \infty } \frac{(-1)^{n+1}}{n}- \sum\limits_{n=1}^{ \infty } \frac{(-1)^{n+1}}{n+1}\tag3$

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

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

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

$\sum\limits_{n=1}^{ \infty } (-1)^{n+1}\frac{2n+1}{n(n+1)}=1\tag5$


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