# 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

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