Prove that $\lim _{x\to \infty \:}(1+\frac{x^x}{x!})^{\frac{1}{x}} = e$

Question

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?

Answer 1

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 $e$

Answer 2

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 $e$.

Answer 3

Using the fact

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

we have

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

Answer 4

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!

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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 $$\lim_{n\to \infty}\frac1n\log\left(1+\frac{n!}{n^n}\right)=0$$