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 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 $\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n^n} \cdot \left(\frac{n^n}{n^n}\right)^{n-1} = \lim_{n\rightarrow \infty} \frac{1}{n^n} \cdot 1 = 0$ And we know that the product has to be greater than or equal to 0, so by squeeze theorem the product approaches 0. Thanks from idontknow 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 . Tags limit, sequence Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post eulid Calculus 1 May 7th, 2016 01:44 PM walter r Calculus 2 April 23rd, 2013 08:39 AM nappysnake Calculus 2 December 11th, 2011 12:17 PM everettjsj2 Calculus 1 February 22nd, 2010 08:20 PM mpire344 Real Analysis 2 March 24th, 2008 02:09 PM

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