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I need to find the exact sum of the following series, $\sum_{n=2}^\infty \frac{(-1)^n}{n^3-n}$.
The solution goes like this:

$\sum_{n=2}^\infty \frac{(-1)^n}{n^3-n}$

$= \frac12\sum_{n=2}^\infty \frac{(-1)^n}{n-1} + \frac12\sum_{n=2}^\infty \frac{(-1)^n}{n+1} +\sum_{n=2}^\infty \frac{(-1)^n}{n}$

$=\frac12\ln2 + \frac12(\ln2 -(1-\frac12)) + \ln2-1$

$= 2\ln2-\frac54$

I understand how to do the first step by partial fractions. But I did not understand the next step. Can you help me understand how these sums become $\ln$s?

user666150
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    Your title doesn't match with the question posted in the context body – Rohan Shinde May 18 '19 at 15:04
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    that the maclaurin series of ln x – Milan May 18 '19 at 15:05
  • Your post will look better and be easier to read is you use $\ln$ instead of $ln$. Similarly for $\sin, \cos$ and so on. – saulspatz May 18 '19 at 15:07
  • I looked at the ln x Maclaurin series, it looks like I can solve this with them. Thanks for the advice about writing ln and. But I am not sure what kind of title would be relevant to the question, so if you have suggestions I can change it. – user666150 May 18 '19 at 15:14
  • Oh I see, I forgot the change the series in the title from my last question, sorry. – user666150 May 18 '19 at 15:21

2 Answers2

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Just use the fact that$$x\in(-1,1]\implies\log(x+1)=x-\frac{x^2}2-\frac{x^3}3+\frac{x^4}4-\cdots$$

José Carlos Santos
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Hint:

$$\dfrac{\left(-\dfrac13\right)^n}{2n+1}=\dfrac1x\cdot\dfrac{x^{2n+1}}{2n+1}$$

where $x=\dfrac i{\sqrt3}$

$$S=2\sum_{r=0}^\infty\dfrac{x^{2n+1}}{2n+1}=\ln(1+x)-\ln(1-x)=\ln\dfrac{1+x}{1-x}=\ln\dfrac{1+\dfrac i{\sqrt3}}{1-\dfrac i{\sqrt3}}=\ln\dfrac{\sqrt3+i}{\sqrt3-i}$$

As $\dfrac{\sqrt3+i}{\sqrt3-i}=\dfrac{\cot\pi/6+i}{\cot\pi/6-i}=e^{i\pi/3}$ using Intuition behind euler's formula

So the principal value of $S$ will be $\dfrac{i\pi}3$

lab bhattacharjee
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