about a series $\sum_{n \geq1}{\frac{2n^2}{3^n}} $ How is this series ? 

$$\sum_{n \geq1}{\frac{2n^2}{3^n}} ?$$

How I made: 
$$a_{n}=\frac{2n^2}{3^n}$$ and then $$\frac{a_{n+1}}{a_n} \leq 1$$ so the series is convergence . 
Is ok ? 
thanks :) 

Since $2n^2 < 4\cdot 2^n$, the series is dominated by a geometric series, so it is convergent. Moreover, we have:
$$ \forall x:|x|<3,\quad \sum_{n\geq 1}\frac{x^n}{3^n} = \frac{x}{3-x}, $$
$$ \forall x:|x|<3,\quad \sum_{n\geq 1}\frac{n x^n}{3^n} = x\cdot\frac{d}{dx}\left(\frac{x}{3-x}\right) = \frac{3x}{(x-3)^2}$$
$$ \forall x:|x|<3,\quad \sum_{n\geq 1}\frac{n^2 x^n}{3^n} = x\cdot\frac{d}{dx}\left(\frac{3x}{(x-3)^2}\right) = \frac{3x(3+x)}{(3-x)^3}$$
so: 
$$\sum_{n\geq 1}\frac{2n^2}{3^n}=2\cdot\frac{3\cdot 4}{8} = 3.$$
