TITLE: Evaluate $\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}.$
QUESTION [26 upvotes]: Problem
Evaluate  $$\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}.$$
My solution
Notice that $$\lim_{n \to \infty}\frac{n+n^2+n^3+\cdots +n^n}{n^n}=\lim_{n \to \infty}\frac{n(n^n-1)}{(n-1)n^n}=\lim_{n \to \infty}\frac{1-\dfrac{1}{n^n}}{1-\dfrac{1}{n}}=1,$$and

$$\lim_{n \to \infty}\frac{1+2^n+3^n+\cdots+n^n}{n^n}=\frac{e}{e-1}.$$

Hence,\begin{align*}\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}&=\lim_{n \to \infty}\frac{\dfrac{n+n^2+n^3+\cdots +n^n}{n^n}}{\dfrac{1+2^n+3^n+\cdots +n^n}{n^n}}\\&=\frac{\lim\limits_{n \to \infty}\dfrac{n+n^2+n^3+\cdots +n^n}{n^n}}{\lim\limits_{n \to \infty}\dfrac{1+2^n+3^n+\cdots +n^n}{n^n}}\\&=1-\frac{1}{e}.\end{align*}
The solution posted above need to quote an uncommon limit. Is there another more simple and more direct solution?

REPLY [5 votes]: I think the above sandwiching methods are only feasible when you already know the result before tackling the problem. I will present a method starting from scratch.
Note that for a special case of Faulhaber’s formula, the numerator equals
$$S=\sum^n_{k=1}k^n=n^n\sum^n_{k=0}\frac{B^+_k}{k!}\frac{n!n^{1-k}}{(n-k+1)!}$$
It is well known that $$\frac{n!n^a}{(n+a)!}=1+o(\cdots)$$ as $n\to\infty$.
Thus, the required limit $$\frac{S}{n^n}\sim \sum^n_{k=0}\frac{B^+_k}{k!}=\frac1{1-e^{-1}}=\color{RED}{\frac{e}{e-1}}$$ by noting the generating function of Bernoulli numbers $$\frac{t}{1-e^{-t}}=\sum^\infty_{k=0}\frac{B^+_kt^k}{k!}$$
P.S.: I don’t know what the $o(\cdots)$ should be. Anyone who knows it please edit my answer.
Also, I don’t know how to type the symbol for asymptotic in Mathjax, please help me...