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?
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3$\begingroup$ Off the top of my head and without giving the problem any thought at all, a partial fraction decomposition seems promising. $\endgroup$– Xander Henderson ♦Jan 21, 2019 at 2:44
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1$\begingroup$ I checked and yes it's definitely a telescoping series. Just do the partial fraction decomposition. $\endgroup$– Ben WJan 21, 2019 at 2:49
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$\begingroup$ $$(-1)^{n+1}\frac{2n+1}{n(n+1)}=(-1)^{n+1}\frac{n+(n+1)}{n(n+1)}=(-1)^{n+1}\frac1{n+1}-(-1)^n\frac1n$$ $\endgroup$– Mark ViolaJan 21, 2019 at 5:08
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$\begingroup$ See also: Does $\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$ Converges? and Sum of two harmonic alternating series. $\endgroup$– Martin SleziakJan 21, 2019 at 15:23
1 Answer
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:
$\sum\limits_{n=1}^{ \infty } (-1)^{n+1}\frac{2n+1}{n(n+1)}=1\tag5$