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

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

marked as duplicate by Cm7F7Bb, Dario, Namaste, Michael Chernick, Carl SchildkrautApr 20 '17 at 22:37

• What is the answer, if it is not $1$? – robjohn Apr 20 '17 at 14:40
• Indeed, I read too fast. Sorry about that. (I have retracted my vote to close). – Arnaud D. Apr 20 '17 at 14:41
• @ robjohn it was given by my friend and told me answer is not $1$. – DXT Apr 20 '17 at 14:42
• Can we take $n^n$ common here? from both numerator and denominator? – Iti Shree Apr 20 '17 at 15:03
• This supposed duplicate is not the same question. Here it is $\sum k^k$ while the other one is $\sum k^n$. They should be dissociated. – zwim Sep 5 '18 at 20:15

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