# Formulae of the Year 2016

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 .

• – lab bhattacharjee Dec 31 '15 at 14:34
• @labbhattacharjee,It's different my limits.But Thank you – math110 Dec 31 '15 at 14:38
• I see that $\lim A_n=\pi/4$. – kmitov Dec 31 '15 at 14:43
• ... because $A_n$ is a Riemann sum? – GEdgar Dec 31 '15 at 14:59
• Hmm... looks like it is the $6^{th}$ term in an Euler-Maclaurin expansion of the integral. $$\frac{1}{2016} = \frac{B_6}{6!}\left[\frac{d^5}{dx^5}\frac{1}{1+x^2}\right]_0^1$$ – achille hui Dec 31 '15 at 18:21

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
$$\begin{array}{|c|}\hline\mbox{}\\ \ds{\quad% \color{#f00}{\lim_{n \to \infty}{1 \over n^{4}\braces{% 1/24 - n\bracks{n\pars{\pi/4 - A_{n}} - 1/4}}}} = \color{#f00}{2016} \quad} \\ \mbox{}\\ \hline \end{array}$$