Limit of a sequence

Let $(a_n)_{n\in \mathbb{N}}$ be a sequence of real numbers such that the series $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{n^s} ,\; s>0$ converges. Prove that:

$$\lim_{n \rightarrow +\infty} \frac{a_1+a_2+\cdots+a_n}{n^s}=0$$

Solution:



 We are using Abel's summation. For each $n \in \mathbb{N}$ we set $\displaystyle R_n = \sum_{k=n}^{\infty} \frac{a_k}{k^s}$ hence $a_k = k^s (R_k - R_{k+1})$. Because the series converges we have that $R_n \rightarrow 0$ and since $\displaystyle S_n = \sup_{k \geq n} |R_k|$ is finite for all $ n \in \mathbb{N}$ we  deduce that $S_n \rightarrow 0$.

Therefore:

$$\begin{aligned}
\left\lvert \frac{a_1 + \dotsc + a_n}{n^s}\right\rvert
&= n^{-s}\left\lvert \sum_{k=1}^n k^s(R_k - R_{k+1})\right\rvert\\
&= n^{-s}\left\lvert \sum_{k=1}^n k^s R_k - \sum_{k=2}^{n+1} (k-1)^s R_k\right\rvert\\
&= n^{-s}\left\lvert R_1 - n^s R_{n+1} + \sum_{k=2}^n \left(k^s - (k-1)^s\right)R_k\right\rvert\\
&\leq \frac{S_1}{n^s} + S_{n+1} + \sum_{k=2}^n \frac{k^s - (k-1)^s}{n^s}S_k\\
&\leq \frac{S_1}{n^s} + S_{n+1} + S_2\sum_{k=2}^{\lfloor\sqrt{n}\rfloor} \frac{k^s - (k-1)^s}{n^s} + \\
 & \quad \quad \quad \quad \quad \quad \quad +S_{\lfloor\sqrt{n}\rfloor+1}\sum_{k=\lfloor\sqrt{n}\rfloor+1}^n\frac{k^s-(k-1)^s}{n^s}\\
&= \frac{S_1}{n^s} + S_{n+1} + S_2\frac{\lfloor\sqrt{n}\rfloor^s-1}{n^s} + S_{\lfloor\sqrt{n}\rfloor+1} \frac{n^s - \lfloor\sqrt{n}\rfloor^s}{n^s}\\
&\leq\frac{S_1}{n^s} + S_{n+1} + \frac{S_2}{n^{s/2}} + S_{\lfloor\sqrt{n}\rfloor+1}.
\end{aligned}$$

Hence for a given $\epsilon>0$ one can find/choose a $n_\epsilon$ that large such that

$$S_{\left \lfloor \sqrt{n_\epsilon } \right \rfloor}<\frac{\epsilon }{4}, \;\; \frac{S_1}{n^{s/2}_\epsilon}<\frac{\epsilon}{4}$$

This means that  $\displaystyle \left | \frac{a_1+a_2+\cdots+a_n}{n^s} \right |<\epsilon$  for each $n\geq n_\epsilon$ and the exercise comes to an end!