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Let $x_n$ be the unique solution of the equation $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$ Find $\lim_{n \to \infty} x_n$

I think that the limit must be $\frac{1}{2}$, because $\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$ is decreasing and convergent to $e$, while $1+\frac{1}{1!}+\dots+\frac{1}{n!}$ is increasing and convergent to $e$, so $$\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}>1+\frac{1}{1!}+\dots+\frac{1}{n!}$$ which means that $\frac{1}{2}>x_n$. I also know that for $a \in [0,\frac{1}{2}), \left(1+\frac{1}{n}\right)^{n+a}$ is eventually increasing, but I don't know how to get a lower bound for $x_n$ which goes to $\frac{1}{2}$

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  • $\begingroup$ Why do you affirm $\;x_n\;$ is the unique solution to that equation? $\endgroup$ – DonAntonio Feb 11 '18 at 19:24
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    $\begingroup$ Because $\left(1+\frac{1}{n} \right)^{n+x}$ is strictly increasing, and when $x\to \infty$ it goes to $\infty$ and when $x \to 0$ it goes to $\left(1+\frac{1}{n}\right)^n$ which is less than that sum. So the equation has a solution and it is unique, name it $x_n$ $\endgroup$ – AndrewC Feb 11 '18 at 19:26
  • $\begingroup$ Or just take logarithms of both sides to get $n+x =$ const $\endgroup$ – saulspatz Feb 11 '18 at 19:28
  • $\begingroup$ Nice question! +1 The key is to estimate the error if the series for $e$ is truncated at $1/n!$ and if one is familiar with the usual proof for irrationality of $e$ then the error estimate of $1/n\cdot n!$ is available directly from memory. $\endgroup$ – Paramanand Singh Feb 12 '18 at 3:40
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Answer. $x_n\to \dfrac{1}{2}$

Explanation. Taylor series remainder $$ \mathrm{e}=1+\frac{1}{1!}+\cdots+\frac{1}{n!}+\frac{\mathrm{e}^{\xi_n}}{(n+1)!} $$ for some $\xi_n\in(0,1)$. Since $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!},$$ Then $$ n+x_n=\frac{\log\left(\mathrm{e}-\dfrac{\mathrm{e}^{\xi_n}}{(n+1)!}\right)}{\log\left(1+\frac{1}{n}\right)}=\frac{1+\log\left(1-\dfrac{\mathrm{e}^{\xi_n-1}}{(n+1)!}\right)}{\dfrac{1}{n}-\dfrac{1}{2n^2}+{\mathcal O}(n^{-3})}=n\cdot\frac{1+{\mathcal O}\left(\frac{1}{(n+1)!}\right)}{1-\dfrac{1}{2n}+{\mathcal O}(n^{-2})}\\=n+\frac{\dfrac{1}{2}+{\mathcal O}(n^{-1})}{1-\dfrac{1}{2n}+{\mathcal O}(n^{-2})} $$ and hence $$ x_n\to \frac{1}{2} $$

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The limit is indeed $\frac{1}{2}$. Due to Taylor's formula with integral remainder,

$$ \sum_{k=0}^{n}\frac{1}{k!} = e-\int_{0}^{1}\frac{(1-t)^n}{n!}\,e^t\,dt=e\left(1+O\left(\tfrac{1}{(n+1)!}\right)\right)=\exp\left(1+O\left(\tfrac{1}{(n+1)!}\right)\right)\tag{1} $$ while $$ \left(1+\frac{1}{n}\right)^{n+x}=\exp\left[(n+x)\left(\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\right]=\exp\left[1+\frac{x-\frac{1}{2}}{n}+o\left(\frac{1}{n}\right)\right]\tag{2} $$ so by equating the RHSs of $(1)$ and $(2)$ we get $\lim_{n\to +\infty}x_n=\frac{1}{2}$ as expected.

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If $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$ then we have $$\begin{align}x_n&=\frac{\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)}{\ln\left( 1+\frac{1}{n}\right)}-n\\&\sim \frac{\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)}{\left(\frac{1}{n}-\frac{1}{2n^2}\right)}-n \\&= n\left( \frac{\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)}{1-\frac{1}{2n}}-1\right)\\&\sim n\left( \ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)-1+\color{blue}{\frac{1}{2n}}\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)\right)&\to \color{red}{\frac12} \end{align}$$

Given that $$\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)\to 1$$

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The definition of $x_n$ can be directly used to evaluate the limit. To simplify typing let $s_n=\sum_{i=0}^{n}1/i!$. Then we have $$x_n=\frac{\log s_n-n\log(1+n^{-1})}{\log(1+n^{-1})}$$ Since $n\log(1+(1/n))\to 1$ the desired limit is equal to the limit of $$n\log s_n-n^2\log(1+(1/n))$$ and now we need to use Taylor series for $\log(1+x)$ to get $$\lim_{n\to\infty} n^2\log(1+(1/n))-n=-\frac{1}{2}$$ and thus our original limit is equal to the limit of $$\frac{1}{2}+n\log s_n-n$$ The next step needs some gymnastics with inequalities. We have $$e-s_n=\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\dots$$ which is less than $$\frac{1}{n!}\left(\frac{1}{n+1}+\frac{1}{(n+1)^2}+\dots\right)=\frac{1}{n!}\frac{1/(n+1)}{1-1/(n+1)}=\frac{1}{n\cdot n!} $$ and therefore we have $$1<\frac{e}{s_n}<1+\frac{1}{ns_n\cdot n!} $$ and applying logs we can see that $$0<n-n\log s_n<n\log\left(1+\frac{1}{ns_n\cdot n!} \right) \leq \frac{1}{s_n\cdot n!} $$ Since $s_n\to e$ by Squeeze Theorem we can see that $n-n\log s_n\to 0$ and therefore the desired limit is $1/2$.

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  • $\begingroup$ +1 Wow! That's exactly the solution I too eventually found! $\endgroup$ – AndrewC Feb 12 '18 at 6:04

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