Clarification Regarding evaluation of $\lim_{n\rightarrow \infty} n\sin(2\pi e n!)$- NBHM-$2009$ 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 :)

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