As I am beginner I don't have any idea about how to find limits, I want to find the limit as $n\rightarrow \infty$ of $$\lim_{n\to\infty}n\cdot \sin(2\pi e(n!)) $$ please help.
Evaluating the limit $\lim_{n\to\infty}n\cdot \sin(2\pi e(n!))$
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sequences-and-series
limits
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1It's okay to be a beginner. What have you tried so far? Can you include that info? – 2017-01-28
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0@dr mv yes that is. – 2017-01-28
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0@ the count I think about this problem as sin value always finite whatever angle it's have and in product of that is n which is very large so according to me the answer is infinity....is I am correct? – 2017-01-28
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0This is the second duplicate today – 2017-01-28
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0What about the tag "sequences-and-series"? – 2017-01-28
2 Answers
3
Let us start with the well-known formula $e=\lim_{n\to\infty}e_n$, where
$$e_n=\sum_{k=0}^n\frac{1}{k!}$$
We observe that $n!e_n$ is an integer. So, if we consider $R_n=e-e_n=\sum_{k=n+1}^\infty\frac{1}{k!}$, we get :
$$\sin(2\pi en!)=\sin(2\pi n!(e_n+R_n))=\sin(2\pi n! R_n)$$
Now, it's not difficult to prove that :
$$\forall n\in\mathbb{N},\quad\frac{1}{(n+1)!}\le R_n\le\frac{1}{n n!}$$
So $$2\pi n! R_n\sim \frac{2\pi}{n}$$
And finally, since $\sin(t)\sim t$ as $t\to 0$ :
$$\lim_{n\to\infty}n\sin(2\pi en!)=2\pi$$
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$\displaystyle e n!=A_n+\frac{1}{n+1}+\sum_{k=n+2}^\infty \frac{n!}{k!}=A_n+\frac{1}{n+1}+O(n^{-2})$ where $A_n\in \mathbb{N}$.
So $$\lim_{n\to \infty}n\sin (2\pi e n!)=\lim_{n\to \infty}n\sin \left(\frac{2\pi}{n+1}+O(n^{-2})\right)=2\pi$$