Prove difference of summations $=\frac{e^2}{2}$

Question

How do I prove that \begin{align} \lim_{n\rightarrow\infty}\left\{\sum\limits_{k=1}^{n}\sum\limits_{i=1}^{\infty}\frac{k\left(2k/n\right)^i}{n\Gamma\left(i+1\right)}-\sum\limits_{k=1}^{n}\sum\limits_{j=1}^{\infty}\frac{k\left(\left(2k-1\right)/n\right)^j}{n\Gamma\left(j+1\right)}\right\}=\frac{1+e^2}{4}? \end{align} I have no idea how to attack this problem. I have little experience or know-how with double summations.

Answer 1

First evaluate the infinite inner series for fixed $n$ and $k$. For the first one we get $$ \sum_{i=1}^\infty\frac{k}{n\cdot i!}\left(\frac{2k}{n}\right)^i=\frac{k}{n}\left(e^\frac{2k}{n}-1\right),\qquad(1) $$ and for the second one $$ \sum_{j=1}^\infty\frac{k}{n\cdot j!}\left(\frac{2k-1}{n}\right)^j=\frac{k}{n}\left(e^\frac{2k-1}{n}-1\right).\qquad (2) $$ The difference (1)-(2) is $$ \frac{k}{n}\left(e^\frac{2k}{n}-e^\frac{2k-1}{n}\right)=\frac{k}{n}\left(e^\frac{2k}{n}-e^\frac{2k-1}{n}\right)=\frac{\left(1-e^\frac{-1}{n}\right)}{n}\cdot k\sqrt[n]{e^2}^k $$ and its sum is $$ \frac{\left(1-e^\frac{-1}{n}\right)}{n}\sum_{k=1}^nk\sqrt[n]{e^2}^k=\left(1-e^\frac{-1}{n}\right)\frac{(n \sqrt[n]{e^2}-n-1)e^\frac{2(n+1)}{n}+\sqrt[n]{e^2}}{n(1-\sqrt[n]{e^2})^2}\to\frac{1+e^2}{4}. $$ I got a different result, so I threw everything into Mathematica, which, however, confirmed my result... There might be a typo in the question?