# Real-analysis: convergence of sequence and convergence of series

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!

• Your example doesn't work: it is convergent since it is positive and smaller than $\sum_{n\ge 1}n^2/(3/2)^{n/2}<\infty$.. – Paolo Leonetti Dec 26 '15 at 16:30
• but dooesn't the $n^2$ in the series make it diverge? – DeLorean Dec 26 '15 at 16:34
• @user301032 Nope. Actually what is stated is true: the series is always convergent. – Crostul Dec 26 '15 at 16:36
• No, $\sqrt(3/2)^n$ has an exponential growth.. – Paolo Leonetti Dec 26 '15 at 16:36

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

• So applying the root test: $\lim_{n\to\infty}(n^2 a_n^\frac{n}{2})^\frac{1}{n}$ gives us $\lim_{n\to\infty}(n^2 )^\frac{1}{n}$ * $\lim_{n\to\infty}(a_n^\frac{n}{2})^\frac{1}{n}$ which then gives us: $\lim_{n\to\infty}\sqrt{a_n}$ which is less then 1 so the sum must converge? – DeLorean Dec 26 '15 at 16:55
• Yes, exactly. Note that Paolo Leonetti's answer is in a sense better, more basic, since in effect he proves the correctness of the Root Test for this particular case, instead of simply quoting the Root Test. – André Nicolas Dec 26 '15 at 17:01
• Which book did/do you use for real-analysis? – DeLorean Dec 26 '15 at 17:41
• @user301032: Rudin. Probably no longer the best, but familiar! – André Nicolas Dec 26 '15 at 17:53

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.

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.

• Where did you learn this? From what book? I want to be able to understand this :) – DeLorean Dec 26 '15 at 17:04