Limits with sequences

Dec 2015
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169
Earth
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. \(\displaystyle l=\lim_{n\rightarrow \infty } \frac{\sum_{i=1}^{n^2 }i^{-1}}{\ln(n)}\).
 

SDK

Sep 2016
804
545
USA
I agree , \(\displaystyle 0<\displaystyle \underbrace{sinsin...sin}_{N}N <1/N \) or \(\displaystyle 0<l\leq \lim_{N\rightarrow \infty } 1/N=0\).
$ 0<l<0 , l=0. $
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.
 
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Dec 2015
1,084
169
Earth
set N=2p and let \(\displaystyle sin_{N}N =\underbrace{sinsin...sin}_{N}N\) , \(\displaystyle \displaystyle sin_N (N) < sin_{N-2}sin\frac{N}{2}<sin_{N-4}sin\frac{N}{4}<...<sin\frac{2p}{2^p}<\frac{2p}{2^p }.\)
Now we have \(\displaystyle 0<l<\lim_{N\rightarrow \infty } \frac{2N}{2^N }=0\).
 
Dec 2015
1,084
169
Earth
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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