Prove that the if $\sum{a_k}$ converges, then $\sum{a_k^2}$ converges Prove that if $a_k\ge0$ and $\sum{a_k}$ converges, then $\sum{a_k^2}$ also converges.
I am not far at all, new to trying to prove math expressions... anyways
all that I have written is $a{_k^2}\le{a_k}$
It makes sense that if a series is convergent then squaring it would make it converge faster at least as k is getting bigger knowing $\lim\limits_{k\to \infty} a_k=0$ for a convergent series...but how would I show this in general?
 A: Here is a brief answer to a brief question:


*

*Justify that there is an $N$ such that for all $n > N$, $a_n < 1$.

*Ignore the finite sum from $1$ to $N$.

*Use your observation that $a_n^2 < a_n$ and basic comparison to conclude that $\sum a_n^2$ is finite.

A: Since  $a_k\ge0$ and $\sum{a_k}$ converges, $a_k\to0$ and so there is a $K$, for $k>K$, $a_k<1$. Thus
$$
\sum_{k=K}^{\infty}a{_k^2}\leqslant \sum_{k=K}^{\infty}a{_k}<\infty
$$
which means 
$$
\sum_{k=1}^{\infty}a{_k^2}=\sum_{k=1}^{K-1}a{_k^2}+\sum_{k=K}^{\infty}a{_k^2}<\infty
$$
A: Hint: Use limit comparison test (positive terms, otherwise they are both trivial for large $n$).
$$ \lim_{n\to \infty} \frac{a_n^2}{a_n}= \lim_{n\to \infty} a_n = 0 $$
Hint2: Partial sum sequence $w_n$, where $$w_n := \sum_{k=1}^n a^2_k,$$ is convergent as it is clearly increasing and bounded from above as
$$ w_n \leq \left(\sum_{k=1}^n a_k\right)^2,$$
as suggested in comments.
A: Let $A_n = \sup_{m\ge n} \sum_{j=n}^{j=m} (a_j)$ . Let$ B_n= \sup_{m\ge n} \sum_{j=n}^{j=m} (a_n^2).$  Since $a_n \ge 0$ and $\sum a_n$ converges, we have $\lim_{n \to \infty} A_n=0 .$ And for all but finitely many $n$ we have $0\le a_n\le 1$ and hence  $0\le a_n^2\le a_n$ (otherwise $ \sum a_n$ doesn't converge.)  Therefore $0 \le B_n \le A_n$ for all but finitely many $n$, so $\lim_{n \to \infty} B_n=0$. Which is necessary and sufficient for $\sum a_n^2$ to converge. FOOTNOTE: The hypothesis $a_n\ge 0$ is necessary. For example if $a_{2 n}=1/\sqrt n$ and $a_{2 n -1}=-1/\sqrt n$ then $\sum_{n=1}^{\infty} ( a_n)=0$ and $\sum a_n^2$ diverges.
