Series with $\sum a_n$ converges but $\sum n a_n^2$ diverges, and $a_n$ is decreasing Is it possible to have a sequence $a_n \geq 0$ which is decreasing and such that $\sum a_n < \infty$ but $\sum n a_n^2 = \infty$? I have seen Find a sequence $a_n$ so that $\sum |a_n|$ converges but $\sum n |a_n|^2$ diverges where the condition holds but the $a_n$ given is not decreasing.
 A: The answer is no, there are two steps to do: 
1) If $a_n$ is decreasing and $\sum_{n=1}^{\infty} a_n < \infty$, show that $a_n \leq 1/n$ for all sufficiently large $n$. 
2) Conclude that if $a_n$ is decreasing and $\sum_{n=1}^{\infty} a_n < \infty$, then  $\sum_{n=1}^{\infty} na_n^2 < \infty$. 
A: The part 1 of @Michael's answer lies on the following fact: $na_{n}\rightarrow 0$ for decreasing/nonincreasing $(a_{n})$, $a_{n}\geq 0$ and that $\displaystyle\sum a_{n}<\infty$.
Consider $na_{2n}\leq a_{2n}+a_{2n-1}+\cdots+ a_{n+1}$ and also consider $na_{2n+1}$. Finally, consider $(2n+1)a_{2n+1}=2(na_{2n+1})+a_{2n+1}$. Note that we have $a_{n}\rightarrow 0$. So now consider the even subsequence and odd subsequence of $(ka_{k})$. 
A: Inspired by the zhw and RRL comments on Marty's answer, here is an example of a  nonnegative and nonincreasing sequence $\{a_n\}_{n=1}^{\infty}$ such that $\sum_{n=1}^{\infty} a_n < \infty$ but $a_i > \frac{1}{i \log(i)}$ for infinitely many  $i$. 

We design the sequence to be constant over a sequence of successive "frames" of time. For frame $k \in \{1, 2, 3, ...\}$ define: 
$$a_n = \frac{5}{(2^{k^2})\log(2^{k^2})} , \quad \mbox{ for } n \in \{2^{(k-1)^2}+1, ..., 2^{k^2}\}$$
Define $a_1=a_2$. For $k \in \{1, 2, 3, ...\}$ define $T_k = 2^{k^2}-2^{(k-1)^2}$, which is the size of frame $k$. 
By construction, this sequence satisfies $a_n = \frac{5}{n\log(n)} > \frac{1}{n\log(n)}$ for indices $n=2^{k^2}$ (for all $k \in \{1, 2, 3, ...\}$). Also: 
\begin{align}
\sum_{n=2}^{\infty} a_n &= \sum_{k=1}^{\infty}\frac{5T_k}{2^{k^2}\log(2^{k^2})}\\
&= \frac{5}{\log(2)}\sum_{k=1}^{\infty} \frac{(T_k/2^{k^2})}{k^2} < \infty
\end{align}
where the last inequality uses the fact that $\lim_{k\rightarrow\infty} \frac{T_k}{2^{k^2}} = 1$. 

The example can of course be modified to make $\{a_n\}$ strictly decreasing if desired. For example, just define $\{b_n\}$ by $b_n = a_n + 1/n^2$. 
A: No.
If $\sum a_n$
converges and
$a_n$ is decreasing,
then
$a_n < \dfrac1{n \log n}$
for all large enough $n$
since the sum of the latter
diverges.
Then
$na_n^2
\lt \dfrac{n}{n^2 \log^2 n}
= \dfrac{1}{n \log^2 n}
$
and the sum of this converges.
