Convergence of the square of the mean series Let $\{a_n\}$ be a sequence of positive numbers and $b_n=\frac{a_1+a_2+\cdots+a_n}{n}$ for $n=1,2,\cdots$. Assume that $\sum_{n=1}^{\infty}a_n^2$ converges.
Then, does $\sum_{n=1}^{\infty}b_n^2$ converge?
 A: EDIT: It seems that the following answer is incorrect, see the comments below. I'm leaving it for posterity but will delete if commenters think that's the better thing to do.

First, we'll sort $\{a_n\}$ in decreasing order, so it's a decreasing sequence.
If $A = \sum a_n^2$ converges, $\sum Ca_n^2$ also converges for any constant $C$. Then consider the first few partial sums of $B = \sum b_n^2$:
$$\frac{a_1^2}{1} + \frac{(a_1+ a_2)^2}{4} + \frac{(a_1+a_2+a_3)^2}{9} + \frac{(a_1+a_2+a_3+a_4)^2}{16} + \cdots$$
That can be rearranged to:
$$B = a_1^2\left(\frac11 + \frac14 + \frac19 +\frac{1}{16}+ \cdots\right) + a^2_2\left(\frac14+\frac19+\frac{1}{16}+\cdots \right) + a_3^2\left(\frac19 + \frac{1}{16} + \cdots \right) + \\
a_1a_2\left(\frac24+\frac29+\frac{2}{16}+\cdots \right) + (a_1a_3+a_2a_3)\left(\frac29+\frac{2}{16}+\cdots \right) + (a_1a_4+a_2a_4+a_3a_4)\left(\frac{2}{16}+\cdots \right)$$
Now all of the terms in parentheses converge to $\pi^2/6$ or less, so we can treat them as constants, and we can rearrange a bit again and split the sum such that $B=S+T$, with $S$ and $T$ as follows:
$$
\begin{align}
S &= C_1a_1^2 + C_2a^2_2 + C_3a_3^2 + \cdots\\
T &= 2C_2(a_1a_2)+ 2C_3(a_1a_3+a_2a_3) + 2C_4(a_1a_4+a_2a_4+a_3a_4) + \cdots \implies\\
T &< 2C_2a_1(a_2+a_3+\cdots) + 2C_3a_2(a_3+a_4+\cdots) + 2C_4a_3(a_4+\cdots)
\end{align}
$$
While $C_n$ is not a constant, it is another decreasing series; hence $S \le (\pi^2/6)A$ and converges. That leaves us to consider $T$.
We know that $A$ converges. This means the terms of $\{a_n\}$ must decrease at least as fast as one of these:

*

*$n^{-r}$, where $r > 0.5$

*$q^{-n}$, where $q > 1$
Let's examine the small products in $T$. If we have $a_n\sim n^{-r}$, then:
$$a_na_{n+1} \sim n^{-r}(n+1)^{-r} = (n^2+n)^{-r} < (n^{-r})^2 \\
a_na_{n+2} \sim (n^2+2n)^{-r} < ((n+1)^{-r})^2 \dots$$
Hence, the series $a_1(a_2+a_3+a_4+\cdots) < a_1^2+a_2^2+a_3^2+\cdots = A$, and the other similar terms in $T$ converge by the same reasoning. Therefore $T$ converges.
If instead $a_n \sim q^{-n}$, we have:
$$a_1a_2 = q^{-3} < a_i^2, a_1a_3 = q^{-4} \le a_2^2, a_1a_4 = q^{-5} < a^2_2$$
and so forth. This one is tougher; if we were to keeping moving this along, we would see that $a_1(a_2+a_3+a_4+ \cdots) < a_1^2+ 2a_2^2 + 2 a_3^2 + \cdots$. But while larger than $A$, that's still smaller than $2A$, which is good enough. Therefore the other similar terms in $T$ converge, and $T$ itself converges.
In all cases, $B=S+T$ converges, and the proposition is correct.
