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Question is to evaluate $$\lim_{n\rightarrow \infty} n\sin(2\pi e n!)$$

We have $e = 1 + \dfrac1{1!} + \dfrac1{2!} + \dfrac1{3!} + \cdots + \dfrac1{n!} + \dfrac1{(n+1)!} + \dfrac1{(n+2)!} + \cdots$

$$n!e=n!(1 + \dfrac1{1!} + \dfrac1{2!} + \dfrac1{3!} + \cdots + \dfrac1{n!} + \dfrac1{(n+1)!} + \dfrac1{(n+2)!} + \cdots)$$

$$=M+\dfrac1{n+1} + \dfrac1{(n+1)(n+2)} + \dfrac1{(n+1)(n+2)(n+3)} + \cdots$$ for some integer $M$.

Now, for $2\pi e n!$ we have :

$$2\pi e n!=2\pi (M+\dfrac1{n+1} + \dfrac1{(n+1)(n+2)} + \dfrac1{(n+1)(n+2)(n+3)} + \cdots)$$

$$=(2\pi M+\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$

For $\sin(2\pi e n!)$ We have :

$$\sin(2\pi e n!)=\sin(2\pi M+\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$

$$=\sin(\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$

For $n\sin(2\pi e n!)$ we have :

$$n\sin(2\pi e n!)=n\sin(\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$

For large $n$ we would have

$$\sin(\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$

$$=\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots$$

I hope I can say that for large $n$

$$n\sin(2\pi e n!)=n(\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$

$$=\frac{2\pi}{1+\frac{1}{n}}+\dfrac{2\pi}{(1+\frac{1}{n})(n+2)} + \dfrac{2\pi}{(1+\frac{1}{n})(n+2)(n+3)} + \cdots)$$

As $n\rightarrow \infty$ we would have :

$$\frac{2\pi}{1+0}+0+0+0+\dots=2\pi$$

So, $$\lim_{n\rightarrow \infty} n\sin(2\pi e n!)=2\pi$$

I would be thankful if some one can check what i have done is reasonably sufficient....

Thank you :)

  • I can not immediately give you a counter example that $\lim(a_n.b_n)\neq \lim a_n \lim b_n$ for just bounded sequence $a_n$ It has to converge to $0$ for them to be equal... I doubt converging to $0$ also need some extra care... –  Jan 16 '14 at 06:15
  • Looks right to me. – hot_queen Jan 16 '14 at 06:21
  • @hot_queen : Thank you :) –  Jan 16 '14 at 06:23
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    http://math.stackexchange.com/questions/76097/what-is-the-limit-of-n-sin-2-pi-cdot-e-cdot-n-as-n-goes-to-infinity – lab bhattacharjee Jan 16 '14 at 06:26
  • @labbhattacharjee : Duplicate? I do not understand this site... The one who tries hard to show some effort would get a duplicate tag from some one else who has just posted the question with irrelevant arguments!! –  Jan 16 '14 at 06:33
  • @PraphullaKoushik He has not casted a close vote. He has given you a link where there is a solution of the problem. Why this outburst? – clark Jan 16 '14 at 06:37
  • @clark : I see that he has not voted to close this... I have seen questions with good effort being closed as duplicate of some other questions with not so considerable ground work.... It would be so frustrating in that case! I should not have been excited so much!! –  Jan 16 '14 at 06:41
  • @labbhattacharjee : I am extremely sorry for being rude.. Thank you for the link :) –  Jan 16 '14 at 06:42

1 Answers1

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your answer seems ok to me. my back-of-the-envelope calc using Landau's "big-O" was: $$ n \sin (2\pi e n) = n \sin(2\pi E_n +2\pi e_n) = n \sin 2\pi e_n \\ = n \sin \left(\frac{2\pi}{n+1} + O(n^{-2})\right) = 2\pi (1+n^{-1})^{-1} + nO(n^{-2}) \\ = 2\pi + O(n^{-1}) $$

David Holden
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  • I am not actually familiar with Big O notation and all.. Thank you for your interest and time :) –  Jan 16 '14 at 06:30
  • i used to intensely dislike it because it seemed somehow vague. but this prejudice was misplaced, and hindered my mathematical education. actually big-O and little-o is a very useful shorthand which, when correctly used, can always be converted into a rigorous argument using ten times as much paper. see e.g. Hardy's "Pure Mathematics" chapter IV section 89 – David Holden Jan 16 '14 at 06:37
  • Thank you for the reference.. I would look into that :) –  Jan 16 '14 at 06:44
  • Hi, how can you prove that :$ \displaystyle\sum_{k=n+2}^{\infty} \frac{n!}{k!}=O(n^{-2})?$ And where is your $o(\frac{2\pi}{n+1} + O(n^{-2}))? $ Thanks in advance. –  Jan 17 '14 at 21:30