Prove that if $\sum |a_{n}|$ converges then $\sum a_{n}^{2}$ converges My attempt runs as follows.
Since $\sum |a_{n}|$ converges, the series $\sum a_{n}$ converges. Since $a_{n}^{2} \geq 0,$ it suffices to prove that there is an $M > 0$ such that
$$\sum_{1}^{m}a_{n}^{2} < M$$
for $m \geq 1.$ But, since
$\sum a_{n}$ converges, there is a $B > 0$ such that
$$\sum_{1}^{m}a_{n} < B$$
for $m \geq 1,$
so
$$(\sum_{1}^{m}a_{n})^{2} = \sum_{1}^{m}a_{n}^{2} + 2\sum_{j < k}a_{j}a_{k} < B^{2}$$
for $m \geq 1.$
However, the problem is there seems no way to determine whether the sum $\sum_{j < k}a_{j}a_{k}$ is positive or negative?
 A: Using the expansion
$$
\left(\sum_{1}^{m}|a_{n}|\right)^{2}=\sum_{1}^{m}a_{n}^{2} + 2\sum_{j < k}|a_{j}||a_{k}|,
$$
you  just deduce
$$
\left(\sum_{1}^{m}|a_{n}|\right)^{2} \geq \sum_{1}^{m}a_{n}^2
$$ and we are done.
A: Since $\sum |a_n|$ converges, $a_n$ converges to zero, which implies $|a_n| < 1$ for all $n$ large enough.
But then $a_n^2 < |a_n| < 1$, and $(a_n)^2$ converges.
A: Well, you have that $\sum |a_n|$ converges to some $B$, so $(\sum |a_n|)^2 = \sum a_n^2 + 2 \sum_{j < k} |a_j| |a_k| = B^2$. $\sum |a_j||a_k| \geq 0$ so it follows that $\sum a_n^2 \leq B^2$, as desired.
A: If $\sum a_n $ converges then $a_n \rightarrow 0$ this implies for all $\epsilon > 0$ there exists $N$ such that $|a_n| <  \epsilon$ for all $n > N$ precisely $|a_n| < 1$ for all $n > N$ then $a_n^2 < |a_n|$ for all $n>N$Hence by comparison test $\sum a_n^2$ is convergent.
A: Another simpler way is to note that $|a_n| \to 0$, so eventually $|a_n|^2$ is dominated by $|a_n|$ and use the comparison test.
A: hint: $|a_n|^2 < |a_n|,\forall n \geq N$
