# 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.

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$$\sum_{n=1}^\infty\ e^{an}\,n^2 \quad\text{for}\;\;a\in R$$

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.)

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Thats a great answer, thank you!! – Alejandro Dec 29 '12 at 16:08
Your very welcome, Alejandro! – amWhy Dec 29 '12 at 16:11
Well, technically, the divergence for $e^a = 1$ does not follow from the root test. – ronno Dec 29 '12 at 19:40
@ronno True! but for $e^a = 1$ the limit is $\infty$, so it diverges – Alejandro Jan 3 '13 at 2:44
@amWhy: I don't think other tests could help us here. – Babak S. Jul 28 '13 at 8:16

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.

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Hint: Use the ratio test.$\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\$

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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 $a$.

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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.

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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.

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