TITLE: Formulae of the Year 2016
QUESTION [24 upvotes]: Decode the following limits to welcome the new year!
This is my love limits (Created by me). I hope you Love it.
Let $$A_{n}=\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2^2}+\cdots+\dfrac{n}{n^2+n^2}$$
show that 
$$\lim_{n\to\infty}\dfrac{1}{n^4\left\{\dfrac{1}{24}-n\left[n\left(\dfrac{\pi}{4}-A_{n}\right)-\dfrac{1}{4}\right]\right\}}=2016$$
can you create some nice other problem (result is 2016)? Happy New  Year To  Everyone .

REPLY [6 votes]: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
A_{n} & \equiv
\sum_{k = 1}^{n}{n \over n^{2} + k^{2}} =
\Im\sum_{k = 0}^{n - 1}{1 \over k + 1 - n\ic} =
\Im\sum_{k = 0}^{\infty}\pars{{1 \over k + 1 - n\ic} -
{1 \over k + n + 1 - n\ic}}
\\[5mm] & =
\Im\bracks{\Psi\pars{n + 1 - n\ic} - \Psi\pars{1 - n\ic}}\,,\qquad\qquad
\pars{~\Psi\ \mbox{is the}\ Digamma\ Function~}.
\end{align}

\begin{align}
A_{n} & =
\Im\braces{\Psi\pars{\bracks{1 - \ic}n} + {1 \over \pars{1 - \ic}n}
- \Psi\pars{-n\ic} - {1 \over -n\ic}}\quad\pars{~Recursion~}
\\[5mm] & =
-\,{1 \over 2n} +
\Im\braces{\Psi\pars{\bracks{1 - \ic}n} - \Psi\pars{-n\ic}}
\end{align}

The Digamma Function Asymptotic Formula  is given by
\begin{align}
\Psi\pars{z} & \sim
\ln\pars{z} - {1 \over 2z} - \sum_{n = 1}^{\infty}{B_{2n} \over 2n\,z^{2n}} =
\ln\pars{z} - {1 \over 2z} - {1 \over 12 z^{2}} + {1 \over 120z^{4}} -
{1 \over 252z^{6}} + \cdots
\\[5mm] & \pars{~z \to \infty\ \mbox{in}\ \verts{\,\mathrm{arg}\pars{z}} < \pi~}
\,,\qquad B_{k}\ \mbox{is a Bernoulli Number.}
\end{align} 

\begin{align}
\Im\Psi\pars{\bracks{1 - \ic}n} & \sim
-\,{\pi \over 4} - {1 \over 4n} - {1 \over 24n^{2}} +
{1 \over \color{#f00}{2016}\,n^{6}} + \cdots
\\[5mm]
\Im\Psi\pars{-n\ic} & \sim
-\,{\pi \over 2} - {1 \over 2n}  + \cdots
\end{align}

\begin{align}
A_{n} &\ \sim\ {\pi \over 4} - {1 \over 4n} - {1 \over 24n^{2}} + {1 \over \color{#f00}{2016}\,n^{6}} + \cdots
\\[5mm]
n\pars{{\pi \over 4} - A_{n}} &\ \sim\
{1 \over 4} + {1 \over 24n} - {1 \over \color{#f00}{2016}\,n^{5}} + \cdots
\\[5mm]
n\pars{{\pi \over 4} - A_{n}} - {1 \over 4} &\ \sim\
{1 \over 24n} - {1 \over \color{#f00}{2016}\,n^{5}} + \cdots
\\[5mm]
n\bracks{n\pars{{\pi \over 4} - A_{n}} - {1 \over 4}} &\ \sim\
{1 \over 24} - {1 \over \color{#f00}{2016}\,n^{4}} + \cdots
\\[5mm]
{1 \over 24} - n\bracks{n\pars{{\pi \over 4} - A_{n}} - {1 \over 4}} &\ \sim\
{1 \over \color{#f00}{2016}\,n^{4}} + \cdots
\\[5mm]
n^{4}\braces{%
{1 \over 24} - n\bracks{n\pars{{\pi \over 4} - A_{n}} - {1 \over 4}}} &\ \sim\
{1 \over \color{#f00}{2016}} +
\pars{~\mbox{terms of order}\ {1 \over n^{2}}~} 
\end{align}

$$
\begin{array}{|c|}\hline\mbox{}\\
\ds{\quad%
\color{#f00}{\lim_{n \to \infty}{1 \over
n^{4}\braces{%
1/24 - n\bracks{n\pars{\pi/4 - A_{n}} - 1/4}}}} = \color{#f00}{2016}
\quad}
\\ \mbox{}\\ \hline
\end{array}
$$