Convergence of the series Im trying to resolve the next exercise:
$$\sum_{n=1}^\infty\ e^{an}n^2   \text{  ,  }a\in R $$
I dont know in which ranges I should separe the a value for resolving the limit and finding out the convergence. 
 A: $$\sum_{n=1}^\infty\ e^{an}\,n^2   \quad\text{for}\;\;a\in R $$
Hint: Use the root test: 
To determine whether $\;\;\sum_{n=1}^\infty b_n\;\;$converges or diverges, evaluate $\;\;\lim_{n\to \infty}\sqrt[\large n]{|b_n|}.\;\;$  In your series, $\;b_n > 0 \;\;\forall n,\;$ so we can drop the absolute value sign:

$$\text{We use the fact that:}\;\; \lim_{n\to \infty}\sqrt[\large n]{n^2} = 1,$$

$$\lim_{n\to \infty} \sqrt[\large n]{e^{an}n^2} \;=\; \lim_{n\to \infty} \sqrt[\large n]{e^{an}}\cdot \sqrt[\large n]{n^2} \;=\; \lim_{n\to \infty}\sqrt[\large n]{e^{an}}\; = \;e^a$$



*

*For what $\;a\;$ is $\;e^a < 1\;$? (At those values, the given series converges.)

*For what values of $\;a\;$ is $\; e^a \gt 1\;$? (At those values, the series diverges.)
A: If $a\geq 0$, the general term does not converge to $0$ so the series diverges.
If $a<0$, we have $\lim_{n\rightarrow +\infty} n^2 (e^{an}n^2)=0$ so there exists a constant $M>0$ such that $0\leq e^{an}n^2 \leq M/n^2$ for all $n\geq 1$.
By comparison, it follows that the series converges.
A: Hint: Use the ratio test.$\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\$
A: Or you can use the root test, knowing that $n^{2/n}\to1$ when $n\to\infty$: $$\lim_{n\to\infty}\sqrt[n]{|u_n|}=L$$ Note that we should consider the values of $a$.
A: Let 
$$S_m= \sum_{n=1}^m \ e^{an}n^2$$
$$T_m= \sum_{n=1}^m\ e^{an}n$$
$$U_m=\sum_{n=1}^m\ e^{an}$$
Since $U_m$ is geometric we know $$U_m=e^a \frac{1-e^{a(m+1)}}{1-e^a}$$
Also
$$T_m-U_m=\sum_{n=1}^m\ e^{an}(n-1)=\sum_{n=0}^{m-1}\ e^{a(n+1)}n=e^a\sum_{n=0}^{m-1}\ e^{an}n = e^a(T_m -me^{am})$$
Therefore
$$T_m(1-e^a)=U_m-me^{a(m+1)}$$
So
$$T_m=\frac{e^a \frac{1-e^{a(m+1)}}{1-e^a}-me^{a(m+1)}}{1-e^a}$$
$$S_m-2T_m+U_m=\sum_{n=1}^m \ e^{an}(n^2-2n+1)=\sum_{n=1}^m \ e^{an}(n-1)^2$$
$$=\sum_{n=0}^{m-1} \ e^{a(n+1)}n^2=e^a(T_m -m^2e^{am})$$
From here you can get a simple closed form for $S_m$, and then decide both on the convergence/divergence and also find the limit.
A: Write it as
$\sum_{n=1}^\infty\ r^n n^2$
where $r = e^a$ satisfies $0 < r$.
If $r \ge 1$ (i.e., $a \ge 0$), the sum clearly diverges.
If $r < 1$ (i.e., $a < 0$), you can get an explicit formula
for $\sum_{n=1}^m\ r^n n^2$
which will show that the sum converges.
Therefore the sum converges for
$a < 0$ and diverges for $a \ge 0$.
The $e^{an}$ seems like a distraction to hide the true nature
of the problem.
