Finding $\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}$ Finding $$\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}$$
Attempt: $$\lim_{n\rightarrow \infty}\bigg[\frac{1}{n^n}+\frac{2^2}{n^n}+\frac{3^3}{n^n}+\cdots \cdots +\frac{n^n}{n^n}\bigg] = 1$$
because all terms are approaching to zero except last terms
but answer is not $1$ , could some help me to solve it , thanks

Bounding by a geometric series,
$$
\begin{align}
&\frac{n^n}{n^n}+\frac{(n-1)^{n-1}}{n^n}+\frac{(n-2)^{n-2}}{n^n}+\cdots+\frac{1^1}{n^n}\\
&\le1+\frac1n+\frac1{n^2}+\frac1{n^3}+\cdots\\
&=\frac{n}{n-1}
\end{align}
$$
Since the sum is obviously always $\ge1$, and $\le\frac{n}{n-1}$, the Squeeze Theorem says that the limit is $1$.
