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We know that

$$\lim_{n\to\infty}n\sin(2\pi \mathrm en!)=2\pi$$

now :

$$\lim_{n\to\infty}n\sin(2^n \pi \sqrt{e}\mathrm n!)=?$$

I tried:

enter image description here

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    I don't understand your first "we know that..." line. – 2017-01-14
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    @SimpleArt :http://math.stackexchange.com/questions/76097/what-is-the-limit-of-n-sin-2-pi-cdot-e-cdot-n-as-n-goes-to-infinity – 2017-01-14
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    you have to bound the fractional part of the sequence inside sine... – 2017-01-14
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    Why does it have a limit? Doesn't it just oscillate indefinitely – 2017-01-14
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    The photo is not clear. Please typeset it. – 2017-01-14
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    Unless you type in what's in the photo, this is likely to be closed. – 2017-01-14

1 Answers 1

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This is tweaked version of the Christian Blatter's answer on the link you posted(hence community wiki). $$2^{n-1}\sqrt{e}n!=2^nn!\sum_{k=0}^\infty\frac{1}{(2^k)k!}=2^{n-1}n!\left(\sum_{k=0}^{n-1}\frac{1}{(2^k)k!}+\sum_{k=n}^ \infty\frac{1}{(2^k)k!}\right)=m_{n-1}+r_{n-1}$$ With $m_{n-1}\in \Bbb{Z}$ and $$\frac{1}{2n}

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    My method is wrong? – 2017-01-14
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    Why do you look at $2^{n-1}\sqrt e n!$? – 2017-01-14
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    @zhw. So I could remove $2\pi m_n$ as a period. – 2017-01-14
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    @Almot1960 I can't really figure out what's written,presumably I can understand first 3 lines but not what you did after. – 2017-01-14