23e = 2 +12!+13!+14!+...
n! (e) = 2(n!) + n!2!+n!3!+n!4!+....
so lim n→∞, n!e = integer
so limn→∞ nsin(2πe n!) = 0
is it correct ?
8 Answers
23
71Agar third step sahi hai.. tab toh sahi hai.. LOL.
Kyoki wahi ek nai samjh aaya.
1708I don,t have answer.
can anyone Conform that It is right or not.
(bcz I also have a doubt in 3rd step.)
341We 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
341In 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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1708$Thanks hsbhatt Sir for giving a nice answer. and clearing doubt.