# Uncomfortable Series Calculations (not geometric nor telescoping): $\sum\limits_{n=1}^{ \infty } (-1)^{n+1}\frac{2n+1}{n(n+1)}$

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?

• Off the top of my head and without giving the problem any thought at all, a partial fraction decomposition seems promising. Jan 21, 2019 at 2:44
• I checked and yes it's definitely a telescoping series. Just do the partial fraction decomposition. Jan 21, 2019 at 2:49
• $$(-1)^{n+1}\frac{2n+1}{n(n+1)}=(-1)^{n+1}\frac{n+(n+1)}{n(n+1)}=(-1)^{n+1}\frac1{n+1}-(-1)^n\frac1n$$ Jan 21, 2019 at 5:08
• Jan 21, 2019 at 15:23

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:

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