Convergence of $\sum\limits_{n=1}^{\infty} (-1)^{n+1}\frac1n\sum\limits_{k=1}^{n}\frac{1}{k}$ $\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}(\sum_{k=1}^{k=n}\dfrac{1}{k})}{n}$
Is this series convergent? 
If I let $a_n=\dfrac{(-1)^{n+1}(\sum_{k=1}^{k=n}\dfrac{1}{k})}{n}=\dfrac{(-1)^{n+1}(1+\dfrac{1}{2}+...+\dfrac{1}{n})}{n}$
and found $\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{n}{n+1}\dfrac{(1+\dfrac{1}{2}+...+\dfrac{1}{n+1})}{(1+\dfrac{1}{2}+...+\dfrac{1}{n})}=\dfrac{n(1+\dfrac{1}{2}+...+\dfrac{1}{n})+\dfrac{n}{n+1}}{n(1+\dfrac{1}{2}+...+\dfrac{1}{n})+(1+\dfrac{1}{2}+...+\dfrac{1}{n})}$
and since $\dfrac{n}{n+1}<1<(1+\dfrac{1}{2}+...+\dfrac{1}{n})$
$\left|\dfrac{a_{n+1}}{a_n}\right|<1$
But I realised that this is insufficient to say that the series converges 
since $lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$ may still be 1 which then the ratio test is inconclusive.
Any method to determine the convergence or divergence of the series?
 A: The series is convergent. We can indeed use a summation by parts argument. We have, denoting $s_k:=\sum_{i=0}^k(-1)^i$ 
\begin{align*}
\sum_{n=M}^{M+N}(-1)^n\frac{\sum_{j=1}^n\frac 1j}{n}&=\sum_{l=M}^{M+N}s_l\frac{\sum_{j=1}^l\frac 1j}{l}-\sum_{l=M-1}^{M+N-1}s_l\frac{\sum_{j=1}^{l+1}\frac 1j}{l+1},
\end{align*}
hence 
$$\left|\sum_{n=M}^{M+N}(-1)^n\frac{\sum_{j=1}^n\frac 1j}{n}\right|\leqslant \frac{2\log(M+N)}{M+N}+\frac{2\log M}M+\sum_{l=M}^{M+N-1}\left|\frac{\sum_{j=1}^{l+1}\frac 1j}{l+1}-\frac{\sum_{j=1}^{l}\frac 1j}{l}\right|.$$
Since 
$$\left|\frac{\sum_{j=1}^{l+1}\frac 1j}{l+1}-\frac{\sum_{j=1}^{l}\frac 1j}{l}\right|\leqslant 2\log l\left(\frac 1l-\frac 1{l+1}\right)+\frac 1{(l+1)^2}$$
and the series $\sum_{k=1}^\infty\frac{\log k}{k^2}$ is convergent, we are done.
A: Recall that a series $\sum a_n$ is called an Alternating Series if (i) the $a_n$ alternate in sign and (ii) $|a_{n+1}|\le |a_n|$ for all $n$ and (iii) $\lim_{n\to\infty} a_n=0$. Any alternating series converges. 
To show (ii) in our case, we need to show that
$$\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1}}{n+1} \le \frac{1+\frac{1}{2}+\cdots +\frac{1}{n}}{n}.$$ 
After "cross-multiplication" and the obvious cancellation, this comes down to showing that
$$\frac{n}{n+1}\le 1+\frac{1}{2}+\cdots +\frac{1}{n},$$
which is clear.
To prove (iii), maybe use the fact that 
$$1+\frac{1}{2}+\cdots+\frac{1}{n}\lt \int_1^{n+1}\frac{dx}{x}=\log(n+1)$$
and $\lim_{n\to\infty}\frac{\log(n+1)}{n}=0$.
A: Note that
$$\lim_{n\to\infty}\frac{\sum_{k=1}^{k=n}\dfrac{1}{k}}{n}=\lim_{n\to\infty}\frac{1/n}{n+1-n}=0,$$
and that
$$\frac{\sum_{k=1}^{k=n}\dfrac{1}{k}}{n}-\frac{\sum_{k=1}^{k=n+1}\dfrac{1}{k}}{n+1}=\frac{1}{n(n+1)}\sum_{k=1}^{k=n}\dfrac{1}{k}-\frac{1}{(n+1)^2}>0,$$
for all $n$. So $\frac{\sum_{k=1}^{k=n}\dfrac{1}{k}}{n}$ decrease to $0$. 
By "alternating series test" we know the series converge.
