Evaluating $\lim\limits_{n \to \infty} \sqrt[n]{\frac{n!}{\sum\limits_{m=1}^n m^m}}$ Evaluate: $$\lim_{n \to \infty} \sqrt[n]{\frac{n!}{\sum_{m=1}^n m^m}}$$
In case it's hard to read, that is the n-th root.  I don't know how to evaluate this limit or know what the first step is...  I believe that: $$\sum_{m=1}^n m^m$$ doesn't have a closed form so I suppose there must be some identity or theorem that must be applied to this limit.  According to the answer key, the limit evaluates to $\frac{1}{e}$.
 A: Let $a_n= (n! / \sum_{m=1}^n m^m)^{1/n}$.  
Observe that $n^n \le \sum_{m=1}^n m^m \le  n n^n$. 
Then $$(\frac{n!}{n^n})^{1/n} \frac{1}{n^{1/n}} \le a_n \le (\frac{n!}{n^n})^{1/n}.$$ 
Since $n^{1/n}\to 1$, we need to find the limit of    $(n! / n^n)^{1/n}$. Take the logarithm of this expression to obtain the Riemann sum 
$$ \frac{1}{n} \sum_{j=1}^n \ln (j/n) \to \int_0^1 \ln x dx = -1.$$ 
Therefore $a_n \to e^{-1}$. 
A: By Stirling's Formula,
$$
n! \approx \left(\frac{n}{e}\right)^n\cdot \sqrt{2\pi n}
$$
In the denominator, we have
$$
\sum_{m=1}^{n}m^m = n^n + (n-1)^{n-1}+\ldots +2^2+1
$$So,
$$
\lim_{n\to\infty}\left(\frac{n!}{\sum_{m=1}^{n}m^m }\right)^{1/n}= \lim_{n\to\infty}\left(\frac{\left(\frac{n}{e}\right)^n\cdot \sqrt{2\pi n}}{\sum_{m=1}^{n}m^m }\right)^{1/n}
$$
$$
=\lim_{n\to\infty}\frac{\frac{n}{e}\cdot (2\pi n)^{1/(2n)}}{\left(\sum_{m=1}^{n}m^m\right)^{1/n} }
$$
$$
=\frac{1}{e}\lim_{n\to\infty}\frac{(2\pi n)^{1/(2n)}}{\frac{1}{n}\left(\sum_{m=1}^{n}m^m\right)^{1/n} }
$$The numerator clearly approaches $1$, so let's just focus on the sum now.
$$
\lim_{n\to\infty}\frac{1}{n}\left(\sum_{m=1}^{n}(m/n)^m\right)^{1/n} =\lim_{n\to\infty}\frac{1}{n}\left(n^n + (n-1)^{n-1}+\ldots +2^2+1\right)^{1/n}
$$
$$
=\lim_{n\to\infty}\left(1 + \frac{(n-1)^{n-1}}{n^n}+\ldots +\frac{2^2}{n^n}+\frac{1}{n^n}\right)^{1/n}
$$Now use the Squeeze Theorem:
$$
1\leq \lim_{n\to\infty}\left(1 + \frac{(n-1)^{n-1}}{n^n}+\ldots +\frac{2^2}{n^n}+\frac{1}{n^n}\right)^{1/n} \leq \lim_{n\to\infty}\left(1 + \frac{(n-1)^{n-1}}{n^n}\cdot (n-1)\right)^{1/n}
$$
$$
1\leq \lim_{n\to\infty}\left(1 + \frac{(n-1)^{n-1}}{n^n}+\ldots +\frac{2^2}{n^n}+\frac{1}{n^n}\right)^{1/n} \leq \lim_{n\to\infty}\left(1 + \frac{(n-1)^{n}}{n^n}\right)^{1/n}
$$In the above equation, the base approaches $1+1/e$, while the exponent approaches $0$ (one could also do a very tedious LHR calculation). Hence we have
$$
1\leq \lim_{n\to\infty}\left(1 + \frac{(n-1)^{n-1}}{n^n}+\ldots +\frac{2^2}{n^n}+\frac{1}{n^n}\right)^{1/n}\leq 1
$$So, in conclusion, the full limit is $e^{-1}$.
