Again quick questions of infinite series

Question

These are series:

1.$\sum_{n=2}^{1000}\frac{2}{n^3-n}=\sum_{n=2}^{1000}\frac{1}{n-1}-\sum_{n=2}^{1000}\frac{2}{n}+\sum_{n=2}^{1000}\frac{1}{n+1}$.

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

Both is to express the sum in different ways possibly. ¿But it's not the same? Which is correct, or is both is correct. ¿Should I factor to solve it? I don't have access to know it. Can I have help? Thanks.

Does this answer your question? Find $ \sum_{n=2}^{+\infty}$ $\frac{1}{n^{3}-n}$ — Toby Mak, Mar 09 '21 at 07:08
$1$ and $2$ are both correct, but note that number $2$ can be written as a telescoping series: see the link in my other comment. To do it your way, you first have to prove that $\frac{2}{n^3-n}$ is absolutely convergent so that you know it is valid to rearrange the terms. — Toby Mak, Mar 09 '21 at 07:10

score 1 · Accepted Answer · answered Mar 09 '21 at 07:02

1

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

is true for all $k\ge2,k\in I$.

The result immediately follows after a quick partial fraction decomposition

answered Mar 09 '21 at 07:02

DatBoi

4,055

Again quick questions of infinite series

1 Answers1