How should I calculate $\lim_{n\rightarrow \infty} \frac{1^n+2^n+3^n+...+n^n}{n^n}$ [duplicate]

Solution 1:

First we use an observation by @Stan in the comment. Note that as $(1 +\frac{x}{n})^n$ is increasing in $n$ whenever $|x|<n$,

$$ \left(\frac{k}{n}\right)^n = \left(1 + \frac{k-n}{n}\right)^n \le e^{k-n}, $$

(here we assume that $x:= k-n$ is fixed and varies the remaining two $n$'s. This sequence is increasing and tends to $e^{k-n}$, as $|x| = |k-n| < n$. See here). Then we have

$$\begin{split} \frac{1^n + 2^n + \cdots + n^n}{n^n} &= \sum_{k=1} ^n \left(\frac{k}{n}\right)^n \\ &\le \sum_{k=1}^n e^{k-n} \\ &= 1 + e^{-1} + e^{-2} + \cdots e^{1-n} \\ &\le \frac{1}{1-e^{-1}} = \frac{e}{e-1}. \end{split} $$ This implies

$$\limsup_{n\to \infty} \frac{1^n + 2^n + \cdots + n^n}{n^n} \le \frac{e}{e-1}.$$

On the other hand, fix $k$. Then for all $n >k$, we have

$$\begin{split} \frac{1^n + 2^n + \cdots + n^n}{n^n} &\ge \frac{(n-k)^n + (n-k+1)^n + \cdots + n^n} {n^n}\\ &= \left( 1 - \frac kn\right)^n + \left( 1 - \frac {k-1}n\right)^n + \cdots +1 \end{split}$$

Then for all $\epsilon >0$, there is $N\in \mathbb N$ so that

$$ \left| \left( 1 - \frac {j-1}n\right)^n - e^{-(j-1)} \right| < \epsilon$$

whenever $n \ge N$ and for all $j = 1, 2 , \cdots, k+1$ (Note $k$ is fixed, so this $N$ can be found)

In particular, this implies

$$ \frac{1^n + 2^n + \cdots + n^n}{n^n} \ge e^{-k} + e^{-(k-1)} + \cdots + 1 - (k+1) \epsilon. $$

Thus

$$ \liminf_{n\to \infty} \frac{1^n + 2^n + \cdots + n^n}{n^n} \ge e^{-k} + e^{-(k-1)} + \cdots + 1 - (k+1) \epsilon. $$

Now let $\epsilon \to 0$ and then $k \to \infty$, we have

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

This implies

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