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April 16th, 2019, 08:34 AM   #1
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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.
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April 16th, 2019, 09:19 AM   #2
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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)^{n-1} = \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.
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April 16th, 2019, 09:44 AM   #3
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I was not able to notice that .
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