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 :) 
 A: Having  ratio $\le 1$ is not enough for convergence. The simplest example is $1+1+1+\cdots$. There are more subtle examples. (By the way,  $\dfrac{a_2}{a_1}\gt 1$, though this does not matter.) 
We want to show that there is a fixed $b$ with $0\le b\lt 1$ such that for large enough $n$, $\dfrac{a_{n+1}}{a_n}\le b$. It will be enough to show that 
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{1}{3}.$$
A: 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: 
$$\sum_{n\geq 1}\frac{2n^2}{3^n}=2\cdot\frac{3\cdot 4}{8} = 3.$$
A: Not quite. 
You need the limit of the ratio $a_{n+1}/a_n$ to be strictly less than one for the ratio test to apply. In symbols, you need:
$$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1 \, . $$ 
In your example, you have:
$$\lim_{n \to \infty}  \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{(n+1)^2}{3n^2}\right| = \frac{1}{3} < 1 \, . $$
Thus, by the ratio test, you sequence converges. Moreover, one can show that:
$$\lim_{k \to \infty} \left(\sum_{n=0}^k \frac{2n^2}{3^n}\right) = 3 \, . $$
