Evaluating $\lim\limits_{n \to \infty} \sqrt[n]{\frac{n!}{\sum\limits_{m=1}^n m^m}}$

Question

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 $\frac{1}{e}$.

Answer 1

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 $a_n \to e^{-1}$.

Answer 2

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 $e^{-1}$.