How to evaluate $\sum_{k=2}^\infty (11k + 7)(k+1)\frac{M^{2k+1}}{(M+N)^{2k+4}}$ as a integral?

Question

There is a summation like this, $$\sum_{k=2}^\infty (11k + 7)(k+1)\frac{M^{2k+1}}{(M+N)^{2k+4}}$$ There is a condition that $\frac{N}{M}\ll 1$.

How to replace the summation over $k$ to an integral and come up with the answer $\frac{11}{4N}$.

I checked this answer using Mathematica and it is correct but can't figure how to convert it to an integral and perform the calculation.

Thanks,

Answer 1

The limit of the series as $M$ goes to infinity and $N>0$ is $$\lim_{M\to +\infty}\sum_{k=2}^\infty (11k + 7)(k+1)\frac{M^{2k+1}}{(M+N)^{2k+4}}=\dfrac{11}{4N^3}.$$

Let $x=(M/(M+N))^2$ then, as $M$ goes to infinity, $$x=\frac{1}{(1+N/M)^2}=1-\frac{2N}{M}+O(1/M^2)$$ and $$ \begin{align} \sum_{k=2}^\infty (11k + 7)(k+1)\frac{M^{2k+1}}{(M+N)^{2k+4}}&= \frac{x^2}{M^3}\sum_{k=2}^\infty (11k^2 + 18k+7)x^k\\ &\sim \frac{11}{M^3}\sum_{k=0}^\infty k^2x^k= \frac{11}{M^3}\sum_{k=0}^\infty k^2e^{k\ln(x)}\\ &\sim \frac{11}{N^3}\cdot\frac{N}{M}\sum_{k=0}^\infty \left(\frac{kN}{M}\right)^2e^{-\frac{2kN}{M}}\\&\to\frac{11}{N^3}\int_0^{\infty}t^2e^{-2t}\,dt=\frac{11}{4N^3}.\end{align}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\sum_{k = 2}^{\infty} \pars{11k + 7}\pars{k + 1}\,{M^{2k + 1} \over \pars{M + N}^{\, 2k + 4}} = {1 \over M^{3}}\sum_{k = 0}^{\infty} \pars{11k + 29}\pars{k + 3}\xi^{k}}$

$\ds{= {11\sum_{k = 0}^{\infty}k^{2}\xi^{k} + 62\sum_{k = 0}^{\infty}k\xi^{k} + 87\sum_{k = 0}^{\infty}\xi^{k} \over M^{3}}\quad}$ where $\ds{\quad\xi \equiv {1 \over \pars{1 + N/M}^{2}}}$

\begin{align} &\left\{\begin{array}{lcl} \ds{\sum_{k = 2}^{\infty}\xi^{k}} & \ds{=} & \ds{{\xi^{2} \over 1 - \xi} = -1 - \xi + {1 \over 1 - \xi}} \\ \ds{\sum_{k = 2}^{\infty}k\,\xi^{k}} & \ds{=} & \ds{-\xi + {\xi \over \pars{1 - \xi}^{2}} = -\xi - {1 \over 1 - \xi} + {1 \over \pars{1 - \xi}^{2}}} \\ \ds{\sum_{k = 2}^{\infty}k^{2}\,\xi^{k}} & \ds{=} & \ds{-\xi + {\xi \over \pars{1 - \xi}^{2}} + \color{red}{2\xi \over \pars{1 - \xi}^{3}}} \end{array}\right. \end{align}

$\ds{N/M \to 0 \implies \xi \to 1}$ such that the original sum becomes

$$ \stackrel{\mrm{as}\ \xi\ \to\ 1}{\sim}\,\,\, {11 \over M^{3}}\,\color{red}{2 \over \pars{1 - \xi}^{3}} \,\,\,\stackrel{\mrm{as}\ N/M\ \to\ 0}{\sim}\,\,\, {11 \over M^{3}}\,{2 \over \pars{2N/M}^{3}} = \bbx{{11 \over 4N^{3}}} $$