<s>[INST] How should I calculate $\lim_{n\rightarrow \infty} \frac{1^n+2^n+3^n+...+n^n}{n^n}$ How should I calculate the below limit 
$$\lim_{n\rightarrow \infty} \frac{1^n+2^n+3^n+...+n^n}{n^n}$$
I have no idea where to start from.
 [/INST] 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 
$$\lim_{n\to \infty} \frac{1^n + 2^n + \cdots + n^n}{n^n} = \frac{e}{e-1}.$$
</s>