# How to show that $n^{-1}s_n \rightarrow 0$ if $n^{-2}s_{n^{2}} \rightarrow 0$?

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?

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

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.
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.