October 15th, 2019, 07:03 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 826 Thanks: 113 Math Focus: Elementary Math  Limits with sequences
Evaluate: a. $\displaystyle l=\lim_{N\rightarrow \infty } \underbrace{sinsin...sin}_{N}N$. b. $\displaystyle l=\lim_{n\rightarrow \infty } n!^{2n}\prod_{i=1}^{n}i^i $. c. $\displaystyle l=\lim_{n\rightarrow \infty } \frac{\sum_{i=1}^{n^2 }i^{1}}{\ln(n)}$. 
October 15th, 2019, 08:30 AM  #2 
Senior Member Joined: Jun 2019 From: USA Posts: 380 Thanks: 205 
a. is quite obviously zero.

October 15th, 2019, 10:17 AM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 826 Thanks: 113 Math Focus: Elementary Math  
October 15th, 2019, 03:08 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 684 Thanks: 457 Math Focus: Dynamical systems, analytic function theory, numerics  Its true that the limit is zero but its not via comparison to $1/N$. The sequence $\sin \sin \sin \cdots \sin N$ refers to function composition, not a product. So $1 < \sin N < 1$ is the last "obvious" step. For instance, in the second iteration you need to prove that $\frac{1}{2} < \sin \sin N < \frac{1}{2}$ and I don't see any reason that this is obviously true.

October 16th, 2019, 12:26 AM  #5 
Senior Member Joined: Dec 2015 From: Earth Posts: 826 Thanks: 113 Math Focus: Elementary Math 
set N=2p and let $\displaystyle sin_{N}N =\underbrace{sinsin...sin}_{N}N$ , $\displaystyle \displaystyle sin_N (N) < sin_{N2}sin\frac{N}{2}<sin_{N4}sin\frac{N}{4}<...<sin\frac{2p}{2^p}<\frac{2p}{2 ^p }.$ Now we have $\displaystyle 0<l<\lim_{N\rightarrow \infty } \frac{2N}{2^N }=0$. 
October 16th, 2019, 04:07 AM  #6 
Senior Member Joined: Dec 2015 From: Earth Posts: 826 Thanks: 113 Math Focus: Elementary Math  b. $\displaystyle \: \frac{n^n }{n!^{2n} } \rightarrow 0 < \displaystyle n!^{2n}\prod_{i=1}^{n}i^i < e^{n^2 }\frac{n^{2n^2 }}{n!^{2n} }=e^{n^2 } (\frac{n^{n}}{n! })^{2n}\rightarrow 0$.


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