# The limit of general term in a series

I have the following statement -

If $\sum_{1}^{\infty} a_{n}^2$ converge then $\sum_{1}^{\infty} a_{n}^3$ converge.

Well i know this statement is true , but if can someone explain why

$\lim_{n\rightarrow\infty}(a_{n}^2) = 0$ implies that $\lim_{n\rightarrow\infty}(a_{n}) = 0$

(a fact that help to prove this statement) , Thanks!

## 2 Answers

An other way (which is almost the method of @DeepSea but approched differently),

Since $a_n\underset{n\to \infty }{\longrightarrow } 0$, there is a $N$ s.t. $a_n<1$ for all $n\geq N$. Therefore, $$|a_n|^3\leq a_n^2$$ provided $n\geq N$. The claim follow.

One way out is to use: $|a_n|^3 - |a_n|^2 = |a_n|^2(|a_n| - 1) \leq 0$ for $n \geq N_0$, and use comparison test. Observe that $- |a_n| \leq a_n \leq |a_n|$. Thus you prove $|a_n| \to 0$ and use sandwich lemma to conclude. To this end, let $\epsilon > 0$ be given, there is an $N$ such that $n > N \implies |a_n|^2 < \epsilon^2\implies |a_n| < \epsilon \implies |a_n| \to 0$.

• Right, i got this idea just dont get why the two limits are equal. – GeorgeB Jun 20 '16 at 11:50
• you can prove it directly with epsilon argument – DeepSea Jun 20 '16 at 11:51