Proof that $ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = \frac{-1}{2} + \frac{4}{3}\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$

Question

Task

Proof that $ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = -\frac{1}{2} + \frac{4}{3}\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$

About

Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ \sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n} $ there is formula for that. So I want to evaluate in the same way left side. So I compute that

$$ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = \sum_{n=2}^{\infty} \frac{1}{3^n \cdot n (n-1)} - \frac{1}{3^n \cdot n (n+1)} $$ I am trying to transform it to use that formula: $$ \sum_{n=1}^{\infty} \frac{1}{n\cdot p^n} = ln\frac{p}{p-1} $$ but I still defeat. So please tell me, there are better ways to proof that or should I consider to change my field of study?

Answer 1

Hints:

$$n^3 - n = (n-1) n (n+1)$$

$$\frac{1}{n (n+1)} = \frac{1}{n} - \frac{1}{n+1}$$ $$\frac{1}{(n-1) n} = \frac{1}{n-1} - \frac{1}{n}$$ $$\frac{1}{(n-1)(n+1)} = \frac{1/2}{n-1} - \frac{1/2}{n+1}$$ $$\frac{1}{(n-1)n(n+1)} = \frac{1}{(n-1)n} - \frac{1}{(n-1)(n+1)} = \frac{1}{n-1} - \frac{1}{n} - \frac{1/2}{n-1} + \frac{1/2}{n+1} = \frac{1/2}{n-1} - \frac{1}{n} + \frac{1/2}{n+1}$$

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

Then try to combine terms when substituting this into the sum.

There is probably a more elegant way to do the following computations... $$\begin{align}\sum_{n \ge 2} \frac{2}{3^n (n-1) n (n+1)} &= \frac{1}{3^2} + \left(\frac{1}{3^3} - \frac{2}{3^2}\right)\frac{1}{2} + \sum_{n \ge 3} \left(\frac{1}{3^{n-1}} - \frac{2}{3^{n}} + \frac{1}{3^{n+1}}\right) \frac{1}{n} \\ &= \frac{1}{54} + \left(3 - 2 + \frac{1}{3}\right)\sum_{n \ge 3} \frac{1}{3^n n} \\ &= \frac{1}{54} + \frac{4}{3} \sum_{n\ge 3} \frac{1}{3^n n} \\ &= \frac{1}{54} - \frac{4}{3} \left(\frac{1}{3} + \frac{1}{9 \cdot 2} \right) + \frac{4}{3} \sum_{n\ge 1} \frac{1}{3^n n} \\ &= - \frac{1}{2} + \frac{4}{3} \sum_{n\ge 1} \frac{1}{3^n n}. \end{align}$$

Answer 2

hint

$$\frac{2}{n^3-n}=\frac{-2}{n}+\frac{1}{n-1}+\frac{1}{n+1}$$