# Prove that the if $\sum{a_k}$ converges, then $\sum{a_k^2}$ converges [duplicate]

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?

## marked as duplicate by Martin Sleziak, Empty, MathOverview, Mankind, user91500Oct 3 '15 at 12:34

• Your phrasing is very unclear. I am assuming $a^2_k \leq a_k$ is given. If so then $$\sum a^2_{k}$$ Converges, since $a^2_k \leq a_k$. – Aleksandar Sep 18 '15 at 1:32
• The inequality $\sum_i a^2_i\le ( \sum_i a_i)^2$ is key here. – Pax Kivimae Sep 18 '15 at 1:33
• We do not need to be given $a_k^2 \le a_k$. – DanielWainfleet Sep 18 '15 at 2:40
• @MartinSleziak You mean it's a corollary to that one. Certainly not a "duplicate". Did you even smell the coffee today? :) – ir7 Sep 18 '15 at 14:45
• @lr7 I don't drink coffee since 2012. I don't have a problem calling it a duplicate - since this result implies the other one and vice-versa. (For series with positive terms, convergence and absolute convergence is equivalent.) But I agree that there is a very minor difference. So I only made comment and left to other users to decide whether it should be closed as a duplicate. – Martin Sleziak Sep 18 '15 at 14:48

Here is a brief answer to a brief question:

1. Justify that there is an $N$ such that for all $n > N$, $a_n < 1$.
2. Ignore the finite sum from $1$ to $N$.
3. Use your observation that $a_n^2 < a_n$ and basic comparison to conclude that $\sum a_n^2$ is finite.

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.

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 $$\sum_{k=1}^{\infty}a{_k^2}=\sum_{k=1}^{K-1}a{_k^2}+\sum_{k=K}^{\infty}a{_k^2}<\infty$$

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.