# Finding the limit of $x_n=\frac{1+2^2+\cdots+n^n}{n^n}$ [duplicate]

I am given $$x_n=\cfrac{1+2^2+\cdots+n^n}{n^n}$$ and asked to find the limit.

I can see that it looks like $$\frac{\displaystyle \sum_{k=1}^n k^k}{n^n}$$ I have been trying to use the squeeze theorem. So far, I have $$1=\cfrac{n^n}{n^n}\leq x_n$$.

• I think it doesn't have a limit. $1/n^n$ diverges Sep 25 '19 at 4:03
• It limits to 1. Split the fraction into two parts. Sep 25 '19 at 4:07
• Sep 25 '19 at 5:00

Squeeze: $$1+2^2 + ... + n^n \leq 1+n^2 + ... + n^n = \frac{n^{n+1}-1}{n-1}$$

Whence the given sequence is dominated by $$\frac{n^{n+1} - 1}{n^n(n-1)} = \frac{n^{n+1} -1}{n^{n+1} - n^n} = 1 + \frac{n^n - 1}{n^{n+1} - n^n} = 1 + \frac{\left(1 - n^{-n}\right)}{n - 1}$$

So the limit is $$1$$.

Hint: try listing terms from right to left. For fixed $$k \ge 0$$

$$\frac{(n-k)^{n-k}}{n^n} \le n^{-k} \ \text{if}\ n \ge k$$

Consider the sequence $$z_n = \frac{1+\cdots + (n-1)^{n-1}}{n^n} = x_n -1$$.

Since $$x^x$$ is monotonic for $$x\geq 1$$, we have

$$\sum_{k=1}^{n-1} k^k \leq \sum_{k=1}^{n-1} n^k < \sum_{k=0}^{n-1} k^k = \frac{n^n-1}{n-1}$$

This gives us the following string of inequalities

$$0 < z_n < \frac{1}{n-1}\left(1-\frac{1}{n^n}\right)$$

so $$z_n \to 0$$ by Squeeze theorem. Thus $$x_n \to 1$$.

As an alternative, by Stolz-Cesaro

$$x_n=\frac{a_n}{b_n}=\cfrac{1+2^2+\cdots+n^n}{n^n}\to 1$$

indeed

$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\cfrac{(n+1)^{n+1}}{(n+1)^{n+1}-n^n}=\frac{1}{ 1-\frac{1}{n+1} \frac{1}{\left(1+\frac 1 n\right)^n} }\to \frac{1}{\frac 1{1-0\cdot \frac 1e}}=1$$