Power Limit!

$\displaystyle \ln(L) = \lim_{n \to \infty}\ln\left({\left(\frac{2n\left(e - \left(1+\frac{1}{n}\right)^n\right)}{e}\right)}^{-n}\right)$

$\displaystyle = \lim_{n \to \infty} -n \ln\left(2n - \frac{2n\left(1 + \frac{1}{n}\right)^n}{e}\right)$

$\displaystyle = \lim_{n \to \infty}-n \ln\left(1 - \left(1 - 2n + \frac{2n\left(1 + \frac{1}{n}\right)^n}{e}\right)\right)$

$\displaystyle = \lim_{n \to \infty} n \displaystyle \sum_{k=1}^{\infty}{\frac{{\left(1 - 2n + \frac{2n\left(1 + \frac{1}{n}\right)^n}{e}\right)}^{k}}{k}}$

$\displaystyle = \lim_{n \to \infty} n\left(1 - 2n + \frac{2n\left(1 + \frac{1}{n}\right)^n}{e}\right) + \frac{n{\left(1 - 2n + \frac{2n\left(1 + \frac{1}{n}\right)^n}{e}\right)}^{2}}{2} + \cdots$

We will consider the Laurent series for $\displaystyle \frac{\left(1 + \frac{1}{n}\right)^n}{e} = 1 - \frac{1}{2n} + \frac{11}{24 n^2} - \frac{7}{16 n^3} + O\left(\frac{1}{n^4}\right)$

We will only consider the first term of our original series and use the value of Laurent series(to a few terms as after a few, denominator will have $n$ and that's unimportant),

$\displaystyle n - 2n^2 + 2n^2\left(1 - \frac{1}{2n} + \frac{11}{24 n^2}\right) = \frac{11}{12}$

$\displaystyle \ln(L) = \frac{11}{12}$

$\displaystyle L = \boxed{{e}^{11/12}}$

$\Big(\frac{2n(e-e_n)}{e}\Big)^{-n}=\Big(\frac{2n(e-(1+n(\frac{1}{n})+\frac{n(n-1)}{2}(\frac{1}{n^2})+\frac{n(n-1)(n-2)}{6}(\frac{1}{n^3})+...))}{e}\Big)^{-n}$ $=\Big(\frac{2n(e-(\sum_{k=0}^n \frac{1}{k!})+\frac{1}{n}(\sum_{k=0}^n \frac{k(k-1)}{2(k!)})-\frac{1}{n^2}(\sum_{k=0}^n \frac{k(k-1)(k-2)(3k-1)}{24(k!)})+\mathcal{O}(\frac{1}{n^3})}{e} \Big)^{-n}$ $=\Big(\frac{2n((\sum_{k=n+1}^\infty \frac{1}{k!})+\frac{1}{n}(\sum_{k=0}^n \frac{k(k-1)}{2(k!)})-\frac{1}{n^2}(\sum_{k=0}^n \frac{k(k-1)(k-2)(3k-1)}{24(k!)})+\mathcal{O}(\frac{1}{n^3})}{e} \Big)^{-n}$ $=\Big(\frac{\mathcal{O}(\frac{1}{n!})+(\sum_{k=0}^n \frac{k(k-1)}{k!})-\frac{1}{n}(\sum_{k=0}^n \frac{k(k-1)(k-2)(3k-1)}{12(k!)})+\mathcal{O}(\frac{1}{n^2})}{e} \Big)^{-n}$ $=\Big(\frac{(\sum_{k=0}^n \frac{1}{k!})-\frac{1}{n}(\sum_{k=0}^n \frac{3k+8}{12(k!)})+\mathcal{O}(\frac{1}{n^2})}{e} \Big)^{-n}$ $=\Big(\frac{(\sum_{k=0}^n \frac{1}{k!})-\frac{1}{n}(\sum_{k=0}^n \frac{1}{4(k!)}+\sum_{k=0}^n \frac{2}{3(k!)})+\mathcal{O}(\frac{1}{n^2})}{e} \Big)^{-n}$ $=\Big(\frac{e-(\sum_{k=n+1}^\infty \frac{1}{k!})-\frac{1}{n}(\sum_{k=0}^n \frac{1}{4(k!)}+\sum_{k=0}^n \frac{2}{3(k!)})+\mathcal{O}(\frac{1}{n^2})}{e} \Big)^{-n}$ $=\Big(\frac{e-\mathcal{O}(\frac{1}{(n+1)!})-\frac{11}{12}\frac{1}{n}\sum_{k=0}^n \frac{1}{k!}+\mathcal{O}(\frac{1}{n^2})}{e} \Big)^{-n}$ $=\Big(\frac{e-\frac{11}{12}\frac{1}{n}(e-\sum_{k=n+1}^\infty \frac{1}{k!})+\mathcal{O}(\frac{1}{n^2})}{e} \Big)^{-n}$ $=\Big(\frac{e-\frac{11}{12}\frac{1}{n}(e)-\mathcal{O}(\frac{1}{(n+2)!})+\mathcal{O}(\frac{1}{n^2})}{e} \Big)^{-n}$ $=\Big(1-\frac{11}{12}\frac{1}{n}+\mathcal{O}(\frac{1}{n^2}) \Big)^{-n}$ ................................................................................................................................................. $\lim_{n\rightarrow \infty} \Big(1-\frac{11}{12}\frac{1}{n}+\mathcal{O}(\frac{1}{n^2}) \Big)^{-n}$ $=\exp(\lim_{n\rightarrow \infty} -n\ln(1-\frac{11}{12}\frac{1}{n}+\mathcal{O}(\frac{1}{n^2})))$ $=\exp\Big(\lim_{n\rightarrow \infty}\frac{\ln(1-\frac{11}{12}\frac{1}{n}+\mathcal{O}(\frac{1}{n^2}))}{\frac{-1}{n}}\Big)$ $=\exp\Big(\lim_{n\rightarrow \infty}\frac{\frac{1}{1-\frac{11}{12}\frac{1}{n}+\mathcal{O}(\frac{1}{n^2})}(\frac{11}{12}\frac{1}{n^2}+\mathcal{O}(\frac{1}{n^3}))}{\frac{1}{n^2}}\Big)$ $=\exp\Big(\lim_{n\rightarrow \infty}\frac{1}{1-\frac{11}{12}\frac{1}{n}+\mathcal{O}(\frac{1}{n^2})}(\frac{11}{12}+\mathcal{O}(\frac{1}{n}))\Big)$ $=e^{11/12}$