Is WolframAlpha wrong about $ \lim_{x\to\infty}(x!)^{1/x}-\frac{x}{e} $? 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.
$$ \lim_{x\to\infty}(x!)^{1/x}-\frac{x}{e} $$
 A: 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 $\log(2\pi x)-1\to\infty$.

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 $3$.
A: 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:
$$\Gamma\left(x+1\right)^{1/x}-\frac{x}{e} = \frac{1}{2e}\ln(2\pi x) + O\left(\frac{1}{x}\right), \quad x\rightarrow \infty$$
A: 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
$$\left(
\begin{array}{cccc}
 k & (1) & (2) & \text{exact} \\
 1 & 0.761595453 & 0.843495044 & 0.849934276 \\
 2 & 1.185132311 & 1.204528536 & 1.204745228 \\
 3 & 1.608669170 & 1.612217034 & 1.612222301 \\
 4 & 2.032206029 & 2.032770401 & 2.032770506
\end{array}
\right)$$
