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

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 05 '18 at 20:15

score 7 · Accepted Answer · answered Apr 20 '17 at 14:45

7

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 $1$.

answered Apr 20 '17 at 14:45

robjohn

353,833

Is this sandwich theorem? – Iti Shree Apr 20 '17 at 15:01
@ItiShree: In the Wikipedia article cited above, the Squeeze Theorem is also known as the Sandwich Theorem and the Pinching Theorem. – robjohn Apr 20 '17 at 17:50

score 5 · Answer 2 · answered Apr 20 '17 at 14:38

5

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

answered Apr 20 '17 at 14:38

Michael Rozenberg

203,855

thanks Michel Rozenberg for nice solution. can we solve it without Stolz method. – DXT Apr 20 '17 at 14:40

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

2 Answers2

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