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