As in title, I want to calculate the following value $$\lim_{n\to \infty}\frac{\sum_{j=1}^n \sum_{k=1}^n k^{1/k^j}}{\sqrt[n]{(\sum_{j=1}^n j!)\sum_{j=1}^n j^n}}.$$
Here is my attempt:
- Since $\sum_{j=1}^n j!\sim n!$, we can replace $\sum_{j=1}^n j!$ by $\sqrt{2\pi n}\frac{n^n}{e^n}$ via Stirling approximation.
- It is clear that $n^n\leq \sum_{j=1}^n n^n$ and $$\frac{1}{n^n}\sum_{j=1}^n j^n=\sum_{j=0}^{n-1} \frac{(n-j)^n}{n^n}=\sum_{j=0}^{n-1}\left(1-\frac{j}{n}\right)^n\leq \sum_{j=0}^\infty e^{-j}=\frac{e}{e-1}.$$ Therefore, $\sum_{j=1}^n j^n\sim \sqrt{C}n^n$ for some constant $C>0$. Then the denominator grows as fast as $$\sqrt[n]{\sqrt{C}n^n\sqrt{2\pi n}\frac{n^n}{e^n}}=\frac{n^2}{e}(\sqrt[2n]{2C\pi n})\sim \frac{n^2}{e}$$ since $\sqrt[2n]{2C\pi n}\to 1$ as $n\to \infty$.
- As for $\sum_{j=1}^n\sum_{k=1}^n k^{1/k^j}$, we first swap the order of summation: $$\sum_{k=1}^n\sum_{j=1}^n k^{1/k^j}.$$ Then for a fixed $k\in \{2,\cdots, n\}$, define the function $f_k:[1,\infty)\to \mathbb R$ by $$f_k(x)=k^{1/k^x}.$$ Thus, we have $$\int_1^n f_k(x)dx\leq \sum_{j=1}^n k^{1/k^j}\leq k^{1/k}+\int_1^n f_k(x)dx$$ since $f_k(x)$ is decreasing. Finally, $$\int_1^n f_k(x)dx=\int_1^n k^{1/k^x}dx\overset{k^{-x}\leftarrow t}{=}\frac{1}{\ln(k)}\int_\frac{1}{k^n}^\frac{1}{k}\frac{1}{t}k^t dt=\frac{1}{\ln(k)}\int_{1/k^n}^{1/k}\frac{1}{t}+O(\ln(k))dt=\frac{1}{\ln(k)}(n\ln(k)-\ln(k)+O(\ln(k)/k))=(n-1)+O(1/k),$$ where the expansion $k^t=1+O(t)$ is from Taylor's expansion: $k^t=1+\sum_{j=1}^\infty \frac{\ln(k)t^j}{j!}=1+O(t)$. Finally, we add up with respect to $k$: $$n+\sum_{k=2}^n(n-1)+O(1/k)\sim n^2.$$
By the discussion above, we have $$\lim_{n\to \infty}\frac{\sum_{j=1}^n \sum_{k=1}^n k^{1/k^j}}{\sqrt[n]{(\sum_{j=1}^n j!)\sum_{j=1}^n j^n}}=\lim_{n\to \infty}\frac{n^2}{n^2/e}=e.$$
I believe the $e$ is indeed the final answer, but I am not sure if there is anything not rigorous enough in my calculations. Thus, I am seeking for a proof verification/correction.
Too long for comments.
You did a good job and your result is correct (very slow convergence by the way).
In fact, the denominator does not make much problems using generalized harmonic number $$\sum_{j=1}^n j^n=H_n^{(-n)}$$ and $$\sum_{j=1}^n j!=\frac{\text{Ei}(1)}{e}-(-1)^n\frac{ \Gamma (n+2)\,\, \Gamma (-n-1,-1)}{e}+\frac{i \pi }{e}-1$$ in place of the subfactorial function.
Concerning the integral, you can go a little bit further since $$\int_{\frac 1{k^n}}^{\frac 1k} \frac{k^t}{t}\,dt=\text{Ei}\left(k^{-1} \log (k)\right)-\text{Ei}\left(k^{-n} \log (k)\right)$$ gives
$$\frac 1{\log(k)}\int_{\frac 1{k^n}}^{\frac 1k} \frac{k^t}{t}\,dt=(n-1)+\frac{1}{k}-k^{-n}+O\left(\frac{1}{\log (k)}\right)$$ and then your result.