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Let $x_n$ be a sequence of real numbers, and put $s_n=x_1+x_2+....+x_n$. Suppose that $n^{-2}s_{n^{2}} \rightarrow 0$ and that the $x_n$ are bounded, and show that $n^{-1}s_{n} \rightarrow 0$.

I am trying to bound $n^{-1}s_{n} \rightarrow 0$ with $n^{-2}s_{n^{2}} \rightarrow 0$ , but not reaching to any conclusion. Can anyone suggest some hints?

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    $\begingroup$ Try to write $n = (\sqrt{n})^2$ and estimate remainder terms using the boundedness of $x_n$. $\endgroup$
    – Zhanxiong
    Oct 7, 2021 at 0:16

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Let $\epsilon > 0$, and let $N$ be large enough so that $\Big| \dfrac{S_{n^2}}{n^2} \Big| < \epsilon$ for all $n \ge N$. Let $|x_n| \le M$, and let $N^2 \le L \le (N + 1)^2 = N^2 + 2N + 1$.

Then, $$L - N^2 \le (N+1)^2 - N^2 = N^2 + 2N + 1 - N^2 = 2N + 1$$

Thus, $$\Big| \dfrac{S_{L}}{L} \Big| = \Big| \dfrac{x_1 + \ldots + x_{N^2} + x_{N^2+1} + \ldots + x_L}{L} \Big| \le \Big| \dfrac{x_1 + \ldots + x_{N^2}}{L} \Big| + \Big| \dfrac{x_{N^2+1} + \ldots + x_L}{L} \Big|$$ $$\le \Big| \dfrac{S_{N^2}}{N^2} \Big| + \dfrac{M \cdot (L - N^2)}{N^2} \le \epsilon + \dfrac{M \cdot (2N + 1)}{N^2} \xrightarrow{N \rightarrow \infty} \epsilon$$

Finally, observe that any natural number, $L \in \mathbb{N}$, sits between two consecutive squares, as above.

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Let $y_n:=s_n/n$. You want to show that $y_n$ is a null sequence and you know that there is a subsequence (the square indices) which converges to $0$. Now try to estimate how big $y_n$ can get between two square numbers.

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