A real analysis problem on convergence Let $x_1,x_2,\dots$ be a sequence of real numbers and put $s_n = x_1 + \dots + x_n$. Suppose that $n^{-2}s_{n^2} \to 0$ and that the $x_n$ are bounded, and show that $n^{-1}s_n \to 0$. 
Tried to show the diff go to 0, but could not get anything. 
Hint enough. 
 A: Hint: (This is an unfinished start to a proof... I'll explain the idea, and why it should work, you can fill in the details.)
We know that $s_{n^{2}}=\frac{1}{n^{2}}\sum_{i=1}^{n^{2}}x_{i}\to 0$.  What we want to do bound how much the terms $s_{n^{2}+k}$ can vary when $1\leq k < 2n+1$; the bound on $k$ insures that $n^{2}<n^{2}+k<(n+1)^{2}$ so that we only inspect the intermediate terms.  Since the $x_{i}$ are bounded, let's say that $x_{i}\leq M$ for all $i$.
Now, note that
\begin{align*}
s_{n^{2}+k} &= \frac{1}{n^{2}+k}\sum_{i=1}^{n^{2}+k}x_{i} \\
            &= \frac{1}{n^{2}+k}\sum_{i=1}^{n^{2}}x_{i} + \frac{1}{n^{2}+k}\sum_{i=n^{2}+1}^{n^{2}+k}x_{i} \\
&\leq \frac{1}{n^{2}+k}\sum_{i=1}^{n^{2}}x_{i} + \frac{kM}{n^{2}+k} \\
&\leq s_{n^{2}} + \frac{kM}{n^{2}} \\
&\leq s_{n^{2}} + \frac{(2n+1)M}{n^{2}}
\end{align*}
Now, note that $s_{n^{2}}\to 0$ and $\frac{(2n+1)M}{n^2}\to 0$.  The latter term can be viewed as an upperbound on the variation from the perfect square terms.  Since the variation goes to $0$, the terms do not vary much from the perfect square terms... hence the whole sequence converges to $0$.
