23
$\begingroup$

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 .

$\endgroup$
  • $\begingroup$ See math.stackexchange.com/questions/469885/… $\endgroup$ – lab bhattacharjee Dec 31 '15 at 14:34
  • $\begingroup$ @labbhattacharjee,It's different my limits.But Thank you $\endgroup$ – math110 Dec 31 '15 at 14:38
  • 1
    $\begingroup$ I see that $\lim A_n=\pi/4$. $\endgroup$ – kmitov Dec 31 '15 at 14:43
  • $\begingroup$ ... because $A_n$ is a Riemann sum? $\endgroup$ – GEdgar Dec 31 '15 at 14:59
  • 6
    $\begingroup$ Hmm... looks like it is the $6^{th}$ term in an Euler-Maclaurin expansion of the integral. $$\frac{1}{2016} = \frac{B_6}{6!}\left[\frac{d^5}{dx^5}\frac{1}{1+x^2}\right]_0^1$$ $\endgroup$ – achille hui Dec 31 '15 at 18:21
5
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.