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October 15th, 2019, 07:03 AM   #1
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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. $\displaystyle l=\lim_{n\rightarrow \infty } \frac{\sum_{i=1}^{n^2 }i^{-1}}{\ln(n)}$.
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October 15th, 2019, 08:30 AM   #2
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a. is quite obviously zero.
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October 15th, 2019, 10:17 AM   #3
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Originally Posted by DarnItJimImAnEngineer View Post
a. is quite obviously zero.
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. $
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October 15th, 2019, 03:08 PM   #4
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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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October 16th, 2019, 12:26 AM   #5
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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 $\displaystyle 0<l<\lim_{N\rightarrow \infty } \frac{2N}{2^N }=0$.
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October 16th, 2019, 04:07 AM   #6
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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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