What is the limit of nsin(2π⋅e⋅n!)n \sin (2 \pi \cdot e \cdot n!) as nn goes to infinity?

I tried and got this

where m is an integer.
\lim_{n\to\infty}n\sin(2\pi en!)=\lim_{n\to\infty}n\sin\left(2\pi n!\sum_{k=0}^n\frac{1}{k!}\right)=\lim_{n\to\infty}n\sin(2\pi m)=\lim_{n\to\infty}n\cdot0=0

Is it correct?


(Added fix recommended by Craig in comments, and complete rewrite for clarity.)

We will use the following: \lim_{x\rightarrow 0} {\frac{\sin x}{x}}=1.

Lemma: If \{x_n\} is a sequence (of non-zero values) that converges to 0, then
\lim_{n\rightarrow\infty}{n \sin{x_n}} = \lim_{n\rightarrow\infty} nx_n

Proof: Rewrite n\sin{x_n} = n x_n \frac{\sin{x_n}}{x_n}. The lemma follows since \sin{x_n}/x_n \rightarrow 1 by above.

Now, let [[x]] be the fractional part of x. Let e_n = [[n!e]].

Lemma: For n>1, e_n\in (\frac{1}{n+1}, \frac{1}{n-1})

n!e = K + \sum_{m=n+1}^\infty \frac{n!}{m!}

Where K is an integer.

But for m>n, \frac{n!}{m!} = \frac{1}{(n+1)(n+2)…m} < n^{n-m}.

So \frac{1}{n+1}<\sum_{m=n+1}^\infty \frac{n!}{m!} < \sum_{m=n+1}^\infty n^{n-m} = \sum_{k=1}^\infty n^{-k}

But the right hand side is a geometric series whose sum is \frac{1}{n-1}.

So n!e-K\in(\frac{1}{n+1}, \frac{1}{n-1}), and, since K is an integer, it must be e_n=n!e-K.

Theorem: \lim_{n\rightarrow \infty} n \sin(2\pi n! e) = 2\pi

Proof: By periodicity of \sin, \sin(2\pi n! e) = \sin(2\pi e_n).

Letting x_n = 2\pi e_n, we see, from our first lemma:

\lim n \sin x_n = \lim n x_n

But nx_n = 2\pi ne_n, and, since ne_n\in(\frac{n}{n+1},\frac{n}{n-1}), we see that ne_n\rightarrow 1. So our limit is 2\pi.

Source : Link , Question Author : M. Amin , Answer Author : Thomas Andrews

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