Prove that $\lim _{x\to \infty \:}(1+\frac{x^x}{x!})^{\frac{1}{x}} = e$ Using a graphing calculator, it seems that $\lim _{x\to \infty \:}(1+\frac{x^x}{x!})^{\frac{1}{x}} = e$. How can this be proven?
 A: Using Stirling, we find:
$$ \frac{x^x}{x!} \asymp \sqrt{2 \pi x} e^x$$
Now the best thing to do would be to write logarithm of your expression and take the limit:
$$\lim_{x \to \infty} \frac{1}{x} \log \left(1+\sqrt{2 \pi x} e^x\right)=\lim_{x \to \infty} \frac{1}{x} \log \left(\sqrt{2 \pi x} e^x\right)=\lim_{x \to \infty} \frac{1}{x} \left( \frac{\log 2}{2}+\frac{\log \pi}{2}+\frac{\log x}{2}+x \right)=1$$
So
$$\lim_{x \to \infty} \log \left(1+\frac{x^x}{x!} \right)^{1/x}=1$$
Meaning, the original limit is $e$
A: Since
$n! > (n/e)^n$,
for integer $x$,
$1+\frac{x^x}{x!}
<1+\frac{x^x}{(x/e)^x}
=1+e^x
$.
Therefore,
for integer $x$,
$(1+\frac{x^x}{x!})^{\frac{1}{x}}
< (1+e^x)^{1/x}
=e (1+e^{-x})^{1/x}
=e (1+\frac{e^{-x}}{x})
=e +e\frac{e^{-x}}{x}
$.
Since
$n! < (n/e)^{n+1}$,
for integer $x$,
$1+\frac{x^x}{x!}
\gt 1+\frac{x^x}{(x/e)^{x+1}}
=1+\frac{e^{x+1}}{x}
$.
Therefore,
for integer $x$,
$(1+\frac{x^x}{x!})^{\frac{1}{x}}
> (1+\frac{e^{x+1}}{x})^{1/x}
= e (e^{-x}+\frac{e}{x})^{1/x}
= e \left(\frac{e}{x}\right)^{1/x}(1+e^{-x}\frac{x}{e})^{1/x}
\gt e +\left(\frac{e}{x}\right)^{1/x}
$.
Therefore,
for integer $x$,
$e \left(\frac{e}{x}\right)^{1/x}
<(1+\frac{x^x}{x!})^{\frac{1}{x}}
<e +e\frac{e^{-x}}{x}
$
and the limits of both sides
are $e$.
A: Using the fact 

$$\lim_{n\to \infty}  a_n^{1/n}=\lim_{n\to \infty} \frac{a_{n+1}}{a_n}  $$

we have

$$\frac{a_{n+1} }{a_n}  =  \frac{ (n+1)! + (1+n)^{1+n} }{(n+1)!+(1+n)n^n} \sim_{n\sim \infty}   \frac{ (1+n)^{1+n} }{n^{n+1} } \longrightarrow_{n\to \infty} e $$

A: We can proceed without appealing to Stirling's Formula.  Rather, we use straightforward arithmetic, elementary inequalities, and evaluation of a Riemann sum.
Note that exploiting the monotonicity of the Gamma function, we can write
$$\lim_{x\to \infty}\left(1+\frac{x^x}{x!}\right)^{1/x}=\lim_{n\to \infty}e^{\frac1n \log\left(1+\frac{n^n}{n!}\right)} $$
Next, we observe that
$$\begin{align}
\lim_{n\to \infty}\frac1n \log\left(1+\frac{n^n}{n!}\right)&=\lim_{n\to \infty}\left(\frac1n\log\left(\frac{n^n}{n!}\right)+\frac1n \log\left(1+\frac{n!}{n^n}\right)\right)\\\\
&\lim_{n\to \infty}\left(-\frac1n\sum_{k=1}^n\log(k/n)+\frac1n \log\left(1+\frac{n!}{n^n}\right)\right) \tag 1\\\\
&=-\int_0^1 \log(x)\,dx \tag 2\\\\
&=1
\end{align}$$ 
Therefore, the continuity of the exponential function guarantees that
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}\left(1+\frac{x^x}{x!}\right)^{1/x}=e}$$
as was to be shown!


In going from $(1)$ to $(2)$ we noted first that $\lim_{n\to \infty}\left(\frac1n\sum_{k=1}^n\log(k/n)\right)$ is the Riemann sum for $\int_0^1 \log(x)\,dx$.  
Then, we exploited the inequalities $\log(1+z)\le z$ for $z>-1$, and $n!\le n^n$, for $n\ge 1$, which shows that $$0\le \frac1n\log\left(1+\frac{n!}{n^n}\right)\le \frac1n$$ whereupon applying the squeeze theorem reveals $$\lim_{n\to \infty}\frac1n\log\left(1+\frac{n!}{n^n}\right)=0$$

