Real-analysis: convergence of sequence and convergence of series

Question

True or false (if true, prove it otherwise give an counterexample). "Suppose $(a_n)$ is a sequence such that $\lim_{n\to\infty}a_n=\alpha$ with $\alpha \in (0,1)$. Then the series $\sum_{n=1}^{\infty}n^2a_n^{n/2}$ is convergent."

I'm guessing this is false. And would use as a counterexample the sequence $a_n=\frac{1}{2}+(\frac{1}{2})^n$. This gives $\lim_{n\to\infty}a_n=\frac{1}{2}$.

But $\sum_{n=1}^{\infty}n^2a_n^{n/2}$=? And this is where I get stuck. I hope someone can help me or give me a hint, thanks in advance!

Answer 1

Hint: Use the Root Test to prove convergence. Note that $\lim_{n\to\infty} (n^2)^{1/n}=1$.

Answer 2

Fix $\varepsilon>0$. Denote by $n_0$ an integer such that $\frac{1}{a_n} \ge \frac{1}{\alpha}-\varepsilon$ whenever $n\ge n_0$. Notice that $\frac{1}{\alpha}>1$. We conclude that $$ \sum_{n\ge 1}n^2a_n^{n/2} \ll \sum_{n\ge n_0} \frac{n^2}{\left(\frac{1}{\alpha}-\varepsilon\right)^{n/2}} < \infty. $$ Therefore it is convergent.

Answer 3

Another attempt.

Since $\lim_n a_n=\alpha \in ]0,1[$, there exists $\beta \in ]0,1[$ such that $\sqrt{\alpha}<\beta<1$. Then $n^2 a_n^{n/2} < \beta^n$ whenever $n\ge n_0$, where $n_0$ is a sufficiently large integer. We conclude that $$ \sum_{n\ge 1}n^2a_n^{n/2} \le \sum_{n=1}^{n_0}n^2a_n^{n/2}+\sum_{n>n_0}\beta^n, $$ which is finite.