Complicated positive series with a harmonic series $$\mathop{\LARGE\mathrm Σ}_{n=1}^\infty \frac{1+1/2+...+1/n}{n}$$
Which are the terms of this sum? I wrote in a paper that the numerator has the terms:
$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$
but I am not sure if this is correct because the series is not the same with:
$\mathop{\LARGE\mathrm Σ}_{n=1}^\infty \frac{1}{n}$.
There is also something else that bothers me,if the denominator is a sum that means you apply the formula for arithmetic progression:$\frac{n(n+1)}{2}$?
I do not understand how to solve this problem because the fraction gives me a lot of confusion.
I tried to solve it with the integral test but I got infinity/infinity(that's because I believed the nominator would be $\frac{1}{n}$ and all the terms added would result in a divergent series).
Can someone explain me?
 A: After Jack D'Aurizio's comment
$$S_p=\sum_{n=1}^{p}\frac{H_n}{n}=\frac{1}{2}\left(H_p\right)^2-\frac{1}{2}\psi ^{(1)}(p+1)+\frac{\pi ^2}{12}$$ Using asymptotics
$$S_p=\frac{1}{2}\log ^2(p)+\gamma  \log (p)+\frac{\pi ^2}{12}+\frac{\gamma ^2}{2}+O\left(\frac{1}{p}\right)$$
As a test, $S_{10}=\frac{32160403}{6350400}\approx 5.06431$ while the above would give $4.96909$.
The relative error will be $\lt 0.1$% as soon as $p > 140$ and  $\lt 0.01$% as soon as $p > 1111.$
A: Summation by parts gives that
$$ \sum_{n=1}^{N}a_n b_n = A_N b_N -\sum_{n=1}^{N-1} A_n (b_{n+1}-b_n) $$
where $A_n=\sum_{k=1}^{n}a_k$. In our case, picking $a_n=\frac{1}{n}$ and $b_n=H_n$ we have
$$\begin{eqnarray*} \sum_{n=1}^{N}\frac{H_n}{n} &=& H_N^2 - \sum_{n=1}^{N-1}\frac{H_n}{n(n+1)}=H_N^2-\sum_{n=1}^{N-1}\frac{H_n}{n}-\sum_{n=1}^{N-1}\frac{H_{n}}{n+1}\\&=&H_N^2-\sum_{n=1}^{N-1}\frac{H_n}{n}-\sum_{n=1}^{N-1}\frac{H_{n+1}}{n+1}+\sum_{n=1}^{N-1}\frac{1}{(n+1)^2}\end{eqnarray*} $$
so
$$ \sum_{n=1}^{N}\frac{H_n}{n} = \frac{H_n^2+H_n^{(2)}}{2}=\frac{1}{2}\log^2N+\gamma\log(N)+\left(\frac{\gamma^2}{2}+\frac{\pi^2}{12}\right)+o(1). $$
A: If your problem is the convergence of the series the it is very easy: since
$$
a_n  = \frac{{1 + \frac{1}
{2} +  \cdots \frac{1}
{n}}}
{n} \geqslant \frac{1}
{n}
$$
for every n and since the series
$$
\sum\limits_{n = 1}^{ + \infty } {\frac{1}
{n}} 
$$
your series is also divergent by direct comparison test. 
A: The terms are $\frac 1 1=1$, $\frac {1+\frac 1 2} {2}=\frac 3  4$, $\frac {1+\frac  12 +\frac 1 3} {3}=\frac {11} {18}$, etc. 
