# $\lim_{n\to\infty}{\left({\sum_{k=1}^nk^k}\right)/{n^n}}=1$?

I'm interested in the following sum $S_n$. $$S_n:=\sum_{k=1}^nk^k=1^1+2^2+3^3+\cdots+n^n.$$

Letting $T_n:={S_n}/{n^n}$, wolfram tells us the followings. $$T_5=1.09216, T_{10}\approx1.04051, T_{30}\approx1.01263, T_{60}\approx1.00622.$$

Then, here is my expectation.

My expectation: $$\lim_{n\to\infty}{T_n}=1.$$

It seems obvious, so I've tried to prove this, but I'm facing difficulty. Then, here is my question.

Question: Could you show me how to find $\lim_{n\to\infty}{T_n}$ if it exists?

Let $n\ge 3$. Look at the top. The sum of the terms up to and including $(n-2)^{n-2}$ is $\le (n-2)(n-2)^{n-2}$. The next term is $(n-1)^{n-1}$ and the last is of course $n^n$.

So our ratio is $\gt 1$ and less than $$\frac{(n-2)(n-2)^{n-2}}{n^n} +\frac{(n-1)^{n-1}}{n^n}+1.\tag{1}$$ The limit of each of the first two terms of (1) is $0$. For the first term is less than $\dfrac{n\cdot n^{n-2}}{n^n}=\dfrac{1}{n}$ and the second is also $\lt \dfrac{1}{n}$.

The result now follows by Squeezing.

• Very nice! I was trying to do something similar, but I did not think of stopping first at $n-2$. Commented Sep 5, 2013 at 8:49

Note that $n\cdot(n-1)^{n-1}\leqslant n^n$ and that $n^2\cdot(n-k)^{n-k}\leqslant n^n$ for every $k\geqslant2$ hence $$n^n\leqslant S_n\leqslant n^n+\frac1n\cdot n^n+(n-2)\cdot\frac1{n^2}\cdot n^n.$$ In particular, for every $n\geqslant1$, $$1\leqslant T_n\leqslant1+\frac2n.$$

We can follow through with the following inequalities:

$$\sum_{k=1}^nk^k=1^1+2^2+\dots (n-1)^{n-1}+n^n<n^n\tag{n>2}$$

$$\sum_{k=1}^nk^k<\sum_{k=1}^nn^k\tag{n>2}=\frac{n^{n+1}-n}{n-1}=\frac{n(n^n-1)}{n-1}=\frac{n^n-1}{1-\frac1n}$$

$$1=\frac{n^n}{n^n}<\frac{\sum_{k=1}^nk^k}{n^n}<\frac{\sum_{k=1}^nn^k}{n^n}=\frac{\frac{n^n-1}{1-\frac1n}}{n^n}=\frac{1-\frac1{n^n}}{1-\frac1n}$$

Taking the limit on both sides should be easy.