How to compute $S_{2016}=\sum\limits_{k=1}^{2016}\left(\sum\limits_{n=k}^{2016}\frac1n\right)^2+\sum\limits_{k=1}^{2016}\frac1k$?
Let $H_n$ be the $n$th Harmonic number with $H_0=0$. $$\begin{align} S_{2016}&=\sum_{k=1}^{2016}\left(\sum_{n=k}^{2016}\frac1n\right)^2+\sum_{k=1}^{2016}\frac1k\\ &=H_{2016}+\sum_{k=1}^{2016} (H_{2016}-H_{k-1})^2\\ &= H_{2016}+\sum_{k=1}^{2016} (H_{2016}^2+H_{k-1}^2-2H_{2016}H_{k-1})\\ &=H_{2016} + 2016H_{2016}^2 -2H_{2016}\sum_{k=1}^{2016}H_{k-1} + \sum_{k=1}^{2016} H_{k-1}^2\\ &=H_{2016} + 2016H_{2016}^2 -2H_{2016}\sum_{k=1}^{2015}H_{k} + \sum_{k=1}^{2015} H_{k}^2 \end{align} $$
As computed here, $\sum_{k=1}^{2015}H_{k}= 2016H_{2015}-2015=2016H_{2016}-2016$ and $$\sum_{k=1}^{2015} H_{k}^2=2016H_{2015}^2-(2\cdot 2016+1)H_{2015} + 2\cdot 2015 $$
Replacing $H_{2015}$ with $H_{2016}-\frac{1}{2016}$ and plugging everything back into the other equality, you should get $S_{2016}=2\cdot 2016$
If you expand all: $$S=1+2\times \frac{1}{2^2}+3\times \frac{1}{3^2}+\cdots +2016 \cdot \frac{1}{2016^2}+$$ $$2\cdot 1\left(1\cdot \frac12+1\cdot \frac13+\cdots 1\cdot \frac{1}{2016}\right)+$$ $$2\cdot 2\left(\frac12 \cdot \frac13+\frac12 \cdot \frac14+\cdots +\frac12 \cdot \frac{1}{2016}\right)+$$ $$\vdots$$ $$2\cdot 2015\left(\frac{1}{2015}\cdot \frac{1}{2016}\right)+$$ $$1+\frac12+\cdots +\frac{1}{2016}=$$ $$2\left(1+\frac12+\cdots+\frac{1}{2016}\right)+$$ $$2\left(\frac12+\frac13+\cdots+\frac{1}{2016}\right)+$$ $$\vdots$$ $$2\left(\frac{1}{2016}\right)=$$ $$2\cdot (2016)=4032.$$
(Please excuse this second solution post - it is a bit too long to be included in the original solution)
$$\scriptsize\begin{align} \color{red}{2n} &=\boxed{\begin{array}{l} &2\big[\;\;\; 1\\ &\;\; +\left(\frac 12+\frac 12\right)\\ &\;\;+\left(\frac 13+\frac 13+\frac 13\right)\\ &\;\;+\qquad \cdots\qquad\ddots\\ &\;\;+\left(\frac 1n+\frac 1n+\frac 1n+\cdots+\frac 1n\right)\big] \end{array}}\\\\ &=\boxed{\begin{array}{r} 2\big[\left(1+\frac 12+\frac 13+\frac 14+\cdots +\frac 1n\right)&\\ +\left(\frac 12+\frac 13+\frac 14+\cdots +\frac 1n\right)&\\ +\left(\frac 13+\frac 14+\cdots +\frac 1n\right)&\\ +\ddots\quad\cdots\quad \vdots\\ +\left(\frac 1n\right)\big] \end{array}}\\\\ &=\boxed{\begin{array}{r} 2\big[1\cdot\frac 11\left(1+\frac 12+\frac 13+\frac 14+\cdots +\frac 1n\right)&\\ +2\cdot \frac 12\left(\frac 12+\frac 13+\frac 14+\cdots +\frac 1n\right)&\\ +3\cdot \frac 13\left(\frac 13+\frac 14+\cdots +\frac 1n\right)&\\ +\ddots\quad\;\cdots\;\quad\vdots\\ +n\cdot \frac 1n\left(\frac 1n\right)\big] \end{array}}\\\\ &=\boxed{\begin{array}{r} 2\big[1\left(1+\frac 12+\frac 13+\frac 14+\cdots+\frac 1n\right)& \\ +\frac 12\left(\frac 12+\frac 13+\frac 14+\cdots+\frac 1n\right)&\\ +\frac 13\left(\frac 13+\frac 14+\cdots+\frac 1n\right)&\\ +\ddots\quad\cdots\quad\vdots\\ +\frac 1n\left(\frac 1n\right)\end{array} \begin{array}{r} \\ +\frac 12\left(\frac 12+\frac 13+\frac 14+\cdots+\frac 1n\right)&\\ +\frac 13\left(\frac 13+\frac 14+\cdots+\frac 1n\right)&\\ +\ddots\quad\cdots\quad\vdots\\ +\frac 1n\left(\frac 1n\right)\end{array} \begin{array}{r} \\ \\ +\frac 13\left(\frac 13+\frac 14+\cdots+\frac 1n\right)&\\ +\ddots\quad\cdots\quad\vdots\\ +\frac 1{n}\left(\frac 1n\right)\end{array} +\cdots \begin{array}{r} \\ \\ \\ \\ +\frac 1n\left(\frac 1n\right)\big]\end{array}}\\\\ &=\boxed{\color{blue}{\begin{array}{r} 2\big[1\left(\quad\frac 12+\frac 13+\frac 14+\cdots+\frac 1n\right)& \\ +\frac 12\left(\quad\frac 13+\frac 14+\cdots+\frac 1n\right)&\\ +\frac 13\left(\quad\frac 14+\cdots+\frac 1n\right)&\\ +\ddots\quad\cdots\quad\vdots\\ +\frac 1{n-1}\left(\quad\frac 1n\right)\end{array}} \color{green}{\begin{array}{r} \\ +\frac 12\left(\quad\frac 13+\frac 14+\cdots+\frac 1n\right)&\\ +\frac 13\left(\quad\frac 14+\cdots+\frac 1n\right)&\\ +\ddots\quad\cdots\quad\vdots\\ +\frac 1{n-1}\left(\quad\frac 1n\right)\end{array}} \color{orange}{\begin{array}{r} \\ \\ +\frac 13\left(\quad\frac 14+\cdots+\frac 1n\right)&\\ +\ddots\quad\cdots\quad\vdots\\ +\frac 1{n-1}\left(\quad\frac 1n\right)\end{array}} +\cdots \begin{array}{r} \\ \\ \\ \\ +\frac 1{n-1}\left(\quad\frac 1n\right)\big]\end{array}\\ + \color{blue}{\begin{array} .\big(1+\frac1{2^2}+\frac 1{3^2}+\frac 1{4^2}+\cdots+\frac 1{n^2}\big)\end{array}} +\color{green}{\begin{array}.\big(\frac1{2^2}+\frac 1{3^2}+\frac 1{4^2}+\cdots+\frac 1{n^2}\big) \end{array}} +\color{orange}{\begin{array}.\big(\frac 1{3^2}+\frac 1{4^2}+\cdots+\frac 1{n^2}\big) \end{array}} +\cdots\qquad\qquad +\begin{array}.\big(\frac 1{n^2}\big)\end{array} \\ + \color{magenta}{\begin{array} .\big(1+\frac1{2^2}+\frac 1{3^2}+\frac 1{4^2}+\cdots+\frac 1{n^2}\big)\end{array} +\begin{array}.\big(\frac1{2^2}+\frac 1{3^2}+\frac 1{4^2}+\cdots+\frac 1{n^2}\big) \end{array} +\begin{array}.\big(\frac 1{3^2}+\frac 1{4^2}+\cdots+\frac 1{n^2}\big) \end{array} +\cdots\qquad\qquad +\begin{array}.\big(\frac 1{n^2}\big)\end{array}}}\\\\ &= \boxed{ \;\;\color{blue}{\begin{array} .\big(1+\frac12+\frac 13+\frac 14+\cdots+\frac 1n\big)^2 \end{array}} +\quad\color{green}{\begin{array} .\big(\frac12+\frac 13+\frac 14+\cdots+\frac 1n\big)^2 \end{array}} +\;\color{orange}{\begin{array} .\big(\frac 13+\frac 14+\cdots+\frac 1n\big)^2 \end{array}} +\cdots+\begin{array} .\big(\frac 1{n-1}+\frac 1n\big)^2 \end{array}+ \tiny\big(\frac 1n\big)^2\\ +\color{magenta}{\begin{array}.\big(1+2\cdot \frac 1{2^2}+3\cdot \frac 1{3^2}+\cdots+n\cdot \frac 1{n^2}\big)\end{array}}}\\\\ &= \boxed{ \;\;\begin{array} .\big(1+\frac12+\frac 13+\frac 14+\cdots+\frac 1n\big)^2 \end{array} +\quad\begin{array} .\big(\frac12+\frac 13+\frac 14+\cdots+\frac 1n\big)^2 \end{array} +\;\begin{array} .\big(\frac 13+\frac 14+\cdots+\frac 1n\big)^2 \end{array} +\cdots+\begin{array} .\big(\frac 1{n-1}+\frac 1n\big)^2 \end{array}+ \tiny\big(\frac 1n\big)^2\\ +\begin{array}.\big(1+\frac 12+\frac 13+\frac 14+\cdots+\frac 1n\big)\end{array}} \end{align}$$
Addendum
The solution can be expressed in a much neater and compact form using summation notation, and also the following notation which I have just conjured up.
Define the Back-Ended Harmonic Series as $$B_{r,n}=B_r=\frac 1r+\frac 1{r+1}+\frac 1{r+2}+\cdots+\frac 1n\qquad (1\le r\le n;\;\; r, n\in \mathbb Z)$$ and the squares of terms of the Back-Ended Harmonic Series as $$C_{r,n}=C_r=\frac 1{r^2}+\frac 1{(r+1)^2}+\frac 1{(r+1)^2}+\cdots+\frac 1{n^2}\qquad (1\le r\le n;\;\; r, n\in \mathbb Z)$$
Using the definitions above, we have the following useful identities: $$\begin{align} B_r&=\frac 1r+B_{r+1}\tag{*}\qquad \text{(recursive definition)}\\ B_r^2&=C_r+2\sum_{r\le s<t\le n} \frac 1{st}\\ &=C_r+2\sum_{s=r}^n\frac 1s\sum_{t=s+1}^n \frac 1t\\ &=C_r+2\sum_{s=r}^n \frac 1s B_{s+1}\tag{**}\\ \end{align}$$
Now back to the question.
$$\begin{align} 2n &=2\sum_{r=1}^n r\cdot \frac 1r\\ &=2\sum_{r=1}^n\sum_{j=1}^r \frac 1r &&\text{(counting }r)\\ &=2\sum_{j=1}^n\sum_{r=j}^n\frac 1r &&\text {(swapping summation order)}\\ &=2\sum_{j=1}^nB_j&&\text{(by definition)}\\ &=2\sum_{j=1}^n j\cdot \frac 1jB_j\\ &=2\sum_{j=1}^n\sum_{k=1}^j \frac 1j B_j&&\text{(counting }j)\\ &=2\sum_{k=1}^n\sum_{j=k}^n\frac 1j B_j &&\text {(swapping summation order)}\\ &=2\sum_{k=1}^n\sum_{j=k}^n \frac 1j\left(\frac 1j+B_{j+1}\right)&&\text{(using *)}\\ &=2\sum_{k=1}^n\sum_{j=k}^n \frac 1jB_{j+1}+\sum_{k=1}^n \sum_{j=k}^n\frac 1{j^2} +\sum_{k=1}^n \sum_{j=k}^n\frac 1{j^2} &&\text{(isolating }\frac 1{j^2})\\ &=\underbrace{\sum_{k=1}^n\left(2\sum_{j=k}^n \frac 1jB_{j+1}+C_k\right)}\;\quad+\sum_{j=1}^n\sum_{k=1}^j \frac 1{j^2} &&\begin{array}.\text{(using }\sum_{j=k}^n \frac 1{j^2}=C_k\\ \text{and swapping summation order)}\end{array}\\ &=\qquad \left(\sum_{k=1}^n B_k^2\right)\;\qquad\qquad+\sum_{j=1}^n j\cdot \frac 1{j^2} &&\text{(using **)}\\ &=\qquad \left(\sum_{k=1}^n B_k^2\right)\;\qquad\qquad+\sum_{j=1}^n \frac 1{j}\\ &=\qquad \left(\sum_{k=1}^n B_k^2\right)\;\qquad\qquad +B_1\\ \end{align}$$