April 16th, 2019, 08:34 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87  Limit with sequence
Evaluate $\displaystyle \lim_{n\rightarrow \infty}\frac{1^1 \cdot 2^2 \cdot 3^3 \cdot ... \cdot n^n }{n^{n^2 }}\; \; $ , $\displaystyle n\in \mathbb{N}$. To write it different : $\displaystyle \lim_{n\rightarrow \infty } n^{n^2 } \prod_{k=1}^{n} k^{k} $. Last edited by idontknow; April 16th, 2019 at 08:36 AM. 
April 16th, 2019, 09:19 AM  #2 
Member Joined: Oct 2018 From: USA Posts: 88 Thanks: 61 Math Focus: Algebraic Geometry 
This may be wrong but i'll take stab at it anyway: $\displaystyle n^{n^2} = (n^{n})^{n}$ So: $\displaystyle \lim_{n\rightarrow \infty}\frac{1^1 \cdot 2^2 \cdot 3^3 \cdot ... \cdot n^n }{n^{n^2 }} =\lim_{n\rightarrow \infty} \frac{1}{n^n} \cdot \frac{2^2}{n^n} ... \frac{n^n}{n^n}$ This is less than $\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n^n} \cdot \left(\frac{n^n}{n^n}\right)^{n1} = \lim_{n\rightarrow \infty} \frac{1}{n^n} \cdot 1 = 0 $ And we know that the product has to be greater than or equal to 0, so by squeeze theorem the product approaches 0. 
April 16th, 2019, 09:44 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87 
I was not able to notice that .


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