Ytukyg

Suppose the sequence ${a_n}_{n=1}^{infty}$ satisfies
$$midsumlimits^{n}_{k=1}{a_{k}}midleq Csqrt{n} spacespacespacespacespacespacespacespacespacespacespacespacespacespacespace n=1, 2, 3, cdots$$
where $C$ is a fixed (but arbitrary) number. Prove that the series
$$sumlimits^{infty}_{n=1}{frac{a_n}{n}}$$
converges.

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

$$<sumlimits^{k-1}_{n=1}{midsumlimits_{i=1}^{n}{a_i}midcdot(b_n-b_{n+1}) +mid sumlimits_{i=1}^{k}{a_i}mid cdot b_k}$$

$$lesumlimits_{n=1}^{k}[Csqrt{n}cdot{(frac{1}{n}-frac{1}{n+1}})]+Csqrt{k}cdot frac{1}{k+1}$$

$$=sumlimits_{n=1}^{k}{frac{Csqrt{n}}{n(n+1)}}+frac{Csqrt{k}}{k+1}.$$

Since $krightarrowinfty$, therefore
$$frac{Csqrt{k}}{k+1}rightarrow 0.$$

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

Above is a $p$-series with $p=frac{3}{2}>1$, hence the series $sumlimits_{n=1}^{k}{frac{Csqrt{n}}{n(n+1)}}$ converges. Even though $sumlimits_{n=1}^{k}{frac{Csqrt{n}}{n(n+1)}}$ converges and $frac{Csqrt{k}}{k+1}$ approaches to $0$ for $krightarrowinfty$, but they do not imply the series $sumlimits^{infty}_{n=1}{frac{a_n}{n}}$ converges.

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

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

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 $sumlimits_{k=1}^nfrac{a_k}k$ is the same as the sequence $sumlimits_{k=1}^nfrac{s_k}{k(k+1)}+frac{s_n}{n+1}$ 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.

In order to prove the convergence, it is enough to show that this sequence is a Cauchy-sequence (another way to prove the convergence is given by robjohn), because $mathbb{R}$, resp. $mathbb{C}$ is complete. Write $A_k := sum_{i=1}^k a_i$ and $B_k := sum_{i=1}^k a_i/i$. Note that $a_k = A_{k} - A_{k-1}$ Then for $n>m$
begin{align}
|B_n - B_m| = left| sum_{i=m+1}^n frac{a_i}{i} right| &= left| sum_{i=m+1}^n frac{A_i -A_{i-1}}{i} right| \
& = frac{|A_m|}{m+1} + frac{|A_n|}{n} + left| sum_{i=m+1}^{n-1} A_i left(frac{1}{i} - frac{1}{i-1} right) right|
end{align}
Since $(i^{-1}-(i-1)^{-1})= -(i(i-1))^{-1}$ Now we can use the bound to get the bound
$$frac{2C}{sqrt{m}}+ C sum_{i=m+1}^{n-1} frac{1}{(i-1) sqrt{i}}.$$
The last sum can be estimated by $int_{m-1}^infty frac{1}{x^{3/2}} d x < frac{3}{2} (m-1)^{-1/2}.$ All in all, we see that $$|B_n-B_m| le frac{3C}{(m-1)^{1/2}},$$
i.e. this is a Cauchy-sequence as claimed.

Note that this argument works also if we only require that $$|sum_{k=1}^n a_k| le C n^{1-varepsilon}.$$
Moreover, we cannot expect that $sum_{k=1}^infty a_k/k$ converges absolutely. Just take $a_k = (-1)^k$, then $$|sum_{k=1}^n a_k| le 1 le sqrt{n},$$ but $$sum_{k=1}^infty frac{|a_k|}{k} = sum_{k=1}^infty frac{1}{k} =infty.$$