# Convergence of $\sum\frac{1}{n^3-n^2}$

Does this infinite series converge or diverge?

$$\sum\frac{1}{n^3-n^2}$$

I've tried every test I can think of but I can't figure it out. Is there a series greater than this that is convergent or a series smaller that is divergent? Or some other quality about this series that I am overlooking?

Please don't tell me to use the integral test, even if that method works.

-
for $n$ large enough you have $n^3-n^2>\frac{1}{2}n^3$ –  drhab Apr 7 at 10:46
Note that $\left(\dfrac{1}{n^3-n^2}\right)_{n \in \mathbb N}\sim _{\infty} \left(\dfrac{1}{n^3}\right)_{n\in \mathbb N}$. –  Git Gud Apr 7 at 10:46
"convergance" in the title should be "convergence". –  Faheem Mitha Apr 7 at 11:21
@FaheemMitha fixed. Presumably that mistake was on purpose because "Convergence of an infinite series" already exists. –  Sabyasachi Apr 7 at 11:46

$0\le \frac{1}{n^3-n^2}=\frac{1}{n^2(n-1)}<\frac {1}{n^2}$ (for all $n>10434666$).

-
Why that extra condition ? n > 10434666 –  Silviu Burcea Apr 7 at 11:26
It's an upper bound, jokingly set way to high, probably. –  5xum Apr 7 at 11:35

Actually, you can sum the series explicitly. $$\sum_{n=2}^\infty \frac{1}{n^3-n^2} = \sum_{n=2}^\infty \frac{1}{n}\left(\frac{1}{n(n-1)}\right) = \sum_{n=2}^\infty \frac{1}{n}\left(\frac{1}{n-1} - \frac{1}{n}\right)\\ = \sum_{n=2}^\infty \left(\frac{1}{n-1}-\frac{1}{n} - \frac{1}{n^2}\right) = 1 - (\zeta(2) - 1) = 2 - \frac{\pi^2}{6}$$

-

For large value of n, the factor behave like 1/n*3, hence by p test it is convergent

-