Limit Question.

I don,t have answer.
can anyone Conform that It is right or not.
(bcz I also have a doubt in 3rd step.)

We have

S = en! = n! \left(1+\frac{1}{2!} + \frac{1}{3!}+...+\frac{1}{n!} + \frac{1}{(n+1)!} +...

N + \left( \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}+... = N+M

where N is an integer. Let us look at M,

M = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}+... > \frac{1}{n+1}

M = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}+... < \frac{1}{n+1} + \frac{1}{(n+1)^2} + \frac{1}{(n+1)^3} +.. < \frac{1}{n}

(Using Geometric Series to infinite terms)

Thus 0<\frac{1}{n+1} < M < \frac{1}{n}<\frac{\pi}{2}

Now \sin (2 \pi en!) = \sin (2 \pi (N+M)) = \sin 2 \pi M

n \sin \frac{2\pi}{n+1} < n \sin (2 \pi M) < n \sin \frac{2\pi}{n}

Now, by Squeeze principle the limit is 2 \pi

In fact this is based on the proof that e is irrational.

Suppose to the contrary, e is rational.

Then e = 1 + \frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{b!} + \frac{1}{(b+1)!}+... = \frac{a}{b} = \frac{(b-1)! a}{b!}

Then b!e is an integer

But from the above we have b!e = an integer + M where

\frac{b}{b+1}<M<1

which clearly is not an integer and thus we have a contradiction.

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$Thanks hsbhatt Sir for giving a nice answer. and clearing doubt.