Given a series converges, prove another similar series converges. Let {$a_n$} be a sequence such that $a_n \ge 0$ for every n in natural numbers and $\sum_{n=1}^{\infty}a_n$ converges. Prove that $\sum_{n=1}^{\infty}a^2_n$ is also convergent.
Proving a series converges seems easy to be because all you have to do is use one of the convergent tests like the ratio test, p-test, etc. How do you use the fact that one series converges to show that a similar series converges? Does it have to do with the fact that I know that because it converges, then there exists a limit and since the series are similar it goes to the same limit and thus converges? I'm a little lost on this problem.
 A: Since $\sum a_n$ converges, we know that $a_n \rightarrow 0$. Therefore, there exists some $N\in \mathbb{N}$ where $$a_n<1, \qquad \forall n>N.$$
Now, if $a_n <1$ we know that $a_n^2<a_n$. By direct comparison test, this suffices to conclude that $\sum a_n^2$ converges.
A: We know that if $\sum a_n$ converges, then $a_n \to 0$. Hence, $a_n$ is bounded, say $a_n \leq M$ for all $n$.
By convergence, there exists $N \in \mathbb N$ such  that $n > N \implies a_n < 1$, then this $\implies a_n^2 < a_n$.
Now, I want you to compare the sequence $a_n^2$ with the sequence :
$$
b_m = \begin{cases}
M^2 \quad m < N \\
a_m \quad m \geq N
\end{cases}
$$
You can see that $b_m \geq a_m^2$ for all $m$, by the way we have chosen $n$. Also, $\sum_{m=1}^\infty b_m = M^2 (N-1) + \sum_{m=N}^\infty a_n$, which is finite.
Hence, by comparison, the square sequence converges.

The important thing here, was to note that if $\sum a_n$ is finite, then $a_n \to 0$. The way one should use convergence of a sequence to prove convergence of other sequences, is by using necessary conditions, such as convergence of partial sums (which I used), convergence of terms to $0$ (again, I used this). Of course, it will come only with practice, but this was a good example.

A: This is quite nice.
First you know that a series converges if and only if its tail converges.
Secondly you know that if a series converges, each of its terms approach $0$.
Therefore consider $N$ such that $n>N \Rightarrow 0 ≤ a_n < 1$. Then $\sum_{i=N}^{\infty}a_n^2 < \sum_{i=N}^{\infty}a_n$, because $a_n^2 < a_n$ for your range of $n$. Can you do the rest?
