Exact sum of the series $\sum_{n=2}^\infty \frac{(-1)^n}{n^3-n}$

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?

• Your title doesn't match with the question posted in the context body – Darkrai May 18 at 15:04
• that the maclaurin series of ln x – Milan May 18 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 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. – Rüzgar Ayan May 18 at 15:14
• Oh I see, I forgot the change the series in the title from my last question, sorry. – Rüzgar Ayan May 18 at 15:21

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

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 $$\dfrac{i\pi}3$$

• I could not understand how to use this in my situation, this is probably too advanced for me, still, thank you for the answer. – Rüzgar Ayan May 18 at 15:20
• @RüzgarAyan, I've followed the problem in the title – lab bhattacharjee May 18 at 15:20
• Yeah it is totally my fault, I am really sorry. – Rüzgar Ayan May 18 at 15:22