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$$\lim_{x\to \infty} (n\cdot \sin(2 \pi e(n!)))$$

I tried a bit using expansion

$$e=1+1+ \frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{n!}$$

$$e*n!= 2n! +\frac{n!}{2!}+\frac{n!}{3!}+\frac{n!}{4!}+\cdots+\frac{n!}{n!}$$

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

I haven't reached till the answer so any help

5xum
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imposter
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    In the first line, did you mean $n$ instead of $x$ in the $x \to \infty$ part? – John Omielan Sep 21 '21 at 06:51
  • I edited your question to improve the mathematical formatting. Please review the changes I made and keep them in mind next time you are writing a question. In particular, you should write $\sin$ instead of $sin$ (because the latter looks like $s$, multiplied by $i$, multiplied by $n$). You should also use $\cdot$ instead of a period to denote multiplication, and $\cdots$ or $\dots$ for ellipsis. – 5xum Sep 21 '21 at 07:00
  • The trick is that when you multiply $e$ by $n!$, the $n$ first terms in the development turn to integer and vanish due to the sine. Continue the expansion. –  Sep 21 '21 at 07:00
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    Finally, and most importantly, you should always write entire equations in math mode, not just the mathematical signs. So, for example, write $e=1+1+ \frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{n!}$ instead of e=1+1+ $\frac{1}{2!}$+$\frac{1}{3!}$+$\frac{1}{4!}$+$\cdots+\frac{1}{n!}$. The former yields $e=1+1+ \frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{n!}$ which looks much better than the latter's "e=1+1+ $\frac{1}{2!}$+$\frac{1}{3!}$+$\frac{1}{4!}$+$\cdots+\frac{1}{n!}$@" – 5xum Sep 21 '21 at 07:02

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