Is WolframAlpha wrong about $ \lim_{x\to\infty}(x!)^{1/x}-\frac{x}{e} $?

Question

I need some help evaluating this limit... Wolfram says it blows up to infinity but I don't think so. I just can't prove it yet.

$$ \lim_{x\to\infty}(x!)^{1/x}-\frac{x}{e} $$

Answer 1

It diverges, albeit slowly.

By Stirling's Approximation, the function approaches

$$f(x)=(2\pi x)^{1/x}\left( \frac xe\right)-\frac xe=\left(\frac xe\right)\left( (2\pi x)^{1/x}-1\right).$$

Now it is true that $(2\pi x)^{1/x}$ goes to $1$ as $x\to\infty$, but we need more information. Using L'Hospital's rule,

\begin{align} \lim_{x\to\infty}\frac{(2\pi x)^{1/x}-1}{e/x}&=\lim_{x\to\infty}\frac{-(2 \pi x)^{1/x} x^{-2} (\log (2 \pi x)-1)}{-ex^{-2}}\\ &=\lim_{x\to\infty}e^{-1}(2 \pi x)^{1/x} (\log (2 \pi x)-1)\\ &=\infty \end{align} since the factor $e^{-1}(2 \pi x)^{1/x}$ is bounded and nonzero and $\log(2\pi x)-1\to\infty$.

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As an aside, check out the harmonic prime series $1/2+1/3+1/5+1/7+\cdots$. This diverges, but so slowly that the first million primes don't even make it past $3$.

Answer 2

I used a simple one-liner and asked Maple 7 to simplify(series(GAMMA(x+1)^(1/x)-x/exp(1), x=infinity,3)); and the answer is 1/2*(ln(2)+ln(Pi)+ln(x)+2*O(1/x)*exp(1))*exp(-1), this can be converted to mathematical notation as: $$\Gamma\left(x+1\right)^{1/x}-\frac{x}{e} = \frac{1}{2e}\ln(2\pi x) + O\left(\frac{1}{x}\right), \quad x\rightarrow \infty$$

Answer 3

In the same spirit as in previous answers, considering $$A=(x!)^{\frac{1}{x}}-\frac{x}{e}=B-\frac{x}{e}$$ $$B=(x!)^{\frac{1}{x}}\implies \log(B)={\frac{1}{x}}\log(x!)$$ Now, using Stirling approximation $$\log(B)={\frac{1}{x}}\left(x (\log (x)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({x}\right)\right)+\frac{1}{12 x}+O\left(\frac{1}{x^3}\right)\right)$$ $$\log(B)= (\log (x)-1)+\frac{1}{2x} \left(\log (2 \pi )+\log \left({x}\right)\right)+O\left(\frac{1}{x^2}\right)$$ Continuing with Taylor $$B=e^{\log(B)}=\frac{x}{e}+\frac{\log (2 \pi )+\log \left({x}\right)}{2 e}+O\left(\frac{1}{x}\right)$$ $$A=\frac{\log (2 \pi x)}{2 e}+O\left(\frac{1}{x}\right)\tag 1$$ Pushing further the expansion of $B$, you would get $$A=\frac{\log (2 \pi x)}{2 e}+\frac{3 \log ^2\left({2 \pi x}\right)+2}{24 e x}+O\left(\frac{1}{x^2}\right)\tag 2$$ For illustration purposes, let $x=10^k$ and computing $$\left( \begin{array}{cccc} k & (1) & (2) & \text{exact} \\ 1 & 0.761595453 & 0.843495044 & 0.849934276 \\ 2 & 1.185132311 & 1.204528536 & 1.204745228 \\ 3 & 1.608669170 & 1.612217034 & 1.612222301 \\ 4 & 2.032206029 & 2.032770401 & 2.032770506 \end{array} \right)$$