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

Question

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 $1=\cfrac{n^n}{n^n}\leq x_n$.

Answer 1

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 $1$.

Answer 2

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

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

Answer 3

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 $x_n \to 1$.

Answer 4

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

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