1. ## convergence help

Suppose that the summation from k=1 to infinity of a_k converges absolutely. Prove that the series summation from k=1 to infinity of (a_k)^2 converges.

2. Originally Posted by friday616
Suppose that the summation from k=1 to infinity of a_k converges absolutely. Prove that the series summation from k=1 to infinity of (a_k)^2 converges.
Problem: Suppose that $\sum_{n=1}^{\infty}a_n$ converges absolutely. Prove that $\sum_{n=1}^{\infty}a^2_n$ converges as well.

Proof: Since, $a_n\to0$ by definition there exists some $N\in\mathbb{N}$ such that $N\le n\implies |a_n|<1$. Therefore, for $N\le n\implies a^2_n=\left|a_n\right|^2<|a_n|$. Therefore $\sum_{n=N}^{\infty}a^2_n<\sum_{n=N}^{\infty}|a_n|$, and since $\sum_{n=1}^{\infty}a_n$ converges absolutely, $\sum_{n=N}^{\infty}|a_n|$ does as well. Therefore, since $\sum_{n=1}^{\infty}a^2_n=\sum_{n=1}^{N}a^2_n+\sum_ {n=N}^{\infty}a^2_n$ both of which converge it is clear that $\sum_{n=1}^{\infty}a^2_n$ converges.

3. You seem to be sulking quantifiers

Also, just a remark, maybe we can discuss about it... Isn't it more logical to put $n\geq N$ instead of $N\leq n$ ? Because we're "conditioning" n, while N is already fixed.

4. Originally Posted by Moo
You seem to be sulking quantifiers

Also, just a remark, maybe we can discuss about it... Isn't it more logical to put $n\geq N$ instead of $N\leq n$ ? Because we're "conditioning" n, while N is already fixed.
Honestly, I think it's semantics. I like the less than or equal sign to be facting left. If you feel that one has a different connotation thatn the other, well then I am sorry.