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I tried and got this

$$e=\sum_{k=0}^\infty\frac{1}{k!}=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}$$ $$n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m$$ 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?

  • 2
    What does the notation ${}^n\sin^{(2\pi en!)}$ mean for you? If you mean $n\sin(2\pi e n!)$, then you might as well just type your calculations into the question as best you can, rather than link to an external rendering that is badly typeset anyway.2011-10-26
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    You go above and beyond, Zev! (re: your edit)2011-10-26
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    Thanks! I figure I'll get that Copy Editor badge eventually :)2011-10-26
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    @Zev: It is certainly a nice fix. But, the $\infty$ in the second equation line should be an $n$.2011-10-26
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    How did you put $n! e=m$ it is not precise!2011-10-26
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    @Joe: Thanks for catching it, I've fixed it now.2011-10-26
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    @Ali: That mistake was my fault.2011-10-26
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    @HenningMakholm I see it typed correctly.. not as you see them. I typed them in Google Docs. Maybe there's something wrong at your end? I don't really know how to use MathJax..2011-10-26
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    This calculation is based on letting $a\to e$ in $n\sin(2\pi a n!)$, while _synchronously_ letting $n\to\infty$. Coalescing limits in this way is not generally valid. At most you can conclude that _if the limit exists_ it will be $0$. But I'm not even sure of that.2011-10-26
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    Ok I can see that @ZevChonoles fixed my post.. thanks Zev.. I also fixed the little mistake about n and infinity.. so.. can anyone check my result tell me is right or wrong?2011-10-26
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    @M.Amin: Didn't you intend for the second line to have an $n$ in the upper limit of the sum, i.e. $$ n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m $$2011-10-26
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    @HenningMakholm: So expressing e itself as a limit inside the bigger limit process is not valid?2011-10-26
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    @Zev: Yes that's how I want it yea..2011-10-26
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    Yes, expressing $e$ as a limit is valid in any context. What is _not_ valid is to combine the two limits and doing them in one operation. Otherwise you could prove $0=1$ by reasoning $$0=\lim_{a\to 0}\;\frac 0a = \lim_{a\to 0}\;\lim_{b\to 0}\;\frac ba = \lim_{x\to 0}\;\frac xx = \lim_{x\to 0}\;1 = 1$$2011-10-26
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    I fixed all the typos now2011-10-26
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    @Henning: But I don't see how lima→0limb→0b/a is equal to limx→0x/x..2011-10-26
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    @M.Amin, it _isn't_. That's the point: collapsing the two limits is not valid, but that is what you were doing in your calculation.2011-10-26
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    @Henning: Ok I get your point. But could you refer me to some text concerning this?2011-10-26
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    @M. Amin: I think you meant $n!\sum_{k=0}^{n}\frac{1}{k!} $ is an integer. is that right?2011-10-26
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    @Gardel: Yes that's what I meant.2011-10-26
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    This might be helpful : fa.its.tudelft.nl/~teuwen/wip/Limit.pdf2014-11-16

3 Answers 3