Finding $\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}$

Question

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

Answer 1

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 $1$.

Answer 2

By Stolz $$\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}=\lim\limits_{n\rightarrow\infty}\frac{n^n}{n^n-(n-1)^{n-1}}=1$$