Show the series $\sum\limits^{\infty}_{n=1}{\frac{a_n}{n}}$ converges

Question

Suppose the sequence $\{a_n\}_{n=1}^{\infty}$ satisfies
$$\mid\sum\limits^{n}_{k=1}{a_{k}}\mid\leq C\sqrt{n} \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space n=1, 2, 3, \cdots$$
where $C$ is a fixed (but arbitrary) number. Prove that the series
$$\sum\limits^{\infty}_{n=1}{\frac{a_n}{n}}$$
converges.

My attempt: Suppose $b_n:= \frac{1}{n}$; Abel's lemma on summation by parts gives
$$\sum\limits^{k}_{n=1}{\frac{a_n}{n}}=\sum\limits^{k-1}_{n=1}{[\sum\limits_{i=1}^{n}{a_i}\cdot(b_n-b_{n+1})] + \sum\limits_{i=1}^{k}{a_i}\cdot b_k}$$

$$<\sum\limits^{k-1}_{n=1}{\mid\sum\limits_{i=1}^{n}{a_i}\mid\cdot(b_n-b_{n+1}) +\mid \sum\limits_{i=1}^{k}{a_i}\mid \cdot b_k}$$

$$\le\sum\limits_{n=1}^{k}[C\sqrt{n}\cdot{(\frac{1}{n}-\frac{1}{n+1}})]+C\sqrt{k}\cdot \frac{1}{k+1}$$

$$=\sum\limits_{n=1}^{k}{\frac{C\sqrt{n}}{n(n+1)}}+\frac{C\sqrt{k}}{k+1}.$$

Since $k\rightarrow\infty$, therefore
$$\frac{C\sqrt{k}}{k+1}\rightarrow 0.$$

Moreover, for the sigma notation, since
$$\frac{C\sqrt{n}}{n(n+1)}<\frac{C\sqrt{n}}{n^2}=\frac{C}{n^\frac{3}{2}}$$

Above is a $p$-series with $p=\frac{3}{2}>1$, hence the series $\sum\limits_{n=1}^{k}{\frac{C\sqrt{n}}{n(n+1)}}$ converges. Even though $\sum\limits_{n=1}^{k}{\frac{C\sqrt{n}}{n(n+1)}}$ converges and $\frac{C\sqrt{k}}{k+1}$ approaches to $0$ for $k\rightarrow\infty$, but they do not imply the series $\sum\limits^{\infty}_{n=1}{\frac{a_n}{n}}$ converges.

My proof seems not yet completed. How do I continue it?

Answer 1

Let $$ s_n=\sum_{k=1}^na_n $$ then $|s_n|\le C\sqrt{n}$ and $$ \begin{align} \sum_{k=1}^n\frac{a_k}k &=\sum_{k=1}^n\frac{s_k-s_{k-1}}k\\ &=\sum_{k=1}^n\frac{s_k}k-\sum_{k=0}^{n-1}\frac{s_k}{k+1}\\ &=\sum_{k=1}^ns_k\left(\frac1k-\frac1{k+1}\right)+\frac{s_n}{n+1}\\ &=\sum_{k=1}^n\underbrace{\frac{s_k}{k(k+1)}}_{\le C\frac{\sqrt{k}}{k^2}}+\underbrace{\ \ \frac{s_n}{n+1}\ \ }_{\le C\frac{\sqrt{n}}{n\vphantom{k^2}}} \end{align} $$ The sum converges and the extra term vanishes.

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Comment brought into the answer

Since it does clarify the answer, I will bring the following comment into the answer.

The sequence of partial sums $\sum\limits_{k=1}^n\frac{a_k}k$ is the same as the sequence $\sum\limits_{k=1}^n\frac{s_k}{k(k+1)}+\frac{s_n}{n+1}$ the latter sum converges absolutely, hence the partial sums are Cauchy and the extra term converges to $0$ hence it is Cauchy. The sum of two Cauchy sequences is Cauchy. Since the original sequence of partial sums is Cauchy, it converges.