How to compute $S_{2016}=\sum\limits_{k=1}^{2016}\left(\sum\limits_{n=k}^{2016}\frac1n\right)^2+\sum\limits_{k=1}^{2016}\frac1k$? I came across a question asking the value of the following sum:
\begin{align}
 \left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\
+\left(\frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\
+ \left(\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\[5 pt]
+\cdots \qquad\quad \vdots\qquad\qquad\\[5 pt]
+ \left(\frac{1}{2015}+\frac{1}{2016}\right)^2 \\
+ \left(\frac{1}{2016}\right)^2\\
+ \left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)\;
\end{align}
I can not find a good way to solve it. Any ideas?

Edit: That is, with no dots, $$S_{2016}=\sum_{k=1}^{2016}\left(\sum_{n=k}^{2016}\frac1n\right)^2+\sum_{k=1}^{2016}\frac1k$$

We show that for any positive integer $n$
$$
S_n:=\sum_{j=1}^n\left(\sum_{k=j}^n\frac{1}{k}\right)^2+\sum_{k=1}^n\frac{1}{k}=\sum_{j=1}^{n} (H_n-H_{j-1})^2 +H_n=2n.$$
where $H_n=\sum_{k=1}^n\frac{1}{k}$.
We have that
\begin{align*}
S_n&=nH_n^2-2H_n\sum_{j=1}^{n}H_{j-1}+\sum_{j=1}^{n}H_{j-1}^2+H_n\\
&=
nH_n^2-2H_n((n+1)H_{n}-n-H_n)\\
&\quad+((n+1)\,H_n^2-(2n+1)\,H_n+2n-H_n^2)+H_n\\
&=2n
\end{align*}
where we used
$$\sum_{j=1}^{n}H_j=(n+1)\,H_n-n,\quad \sum_{j=1}^n H_j^2=(n+1)\,H_n^2-(2n+1)\,H_n+2n$$
(see Sum of Squares of Harmonic Numbers).
