How do we find the following sum (closed form)?
$$S=\sum_{n=1}^{\infty} \frac{\sin ({n})}{n!}$$
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Use (one of) Euler's formulae. – Daniel Fischer Jul 17 '14 at 16:09
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1$\Im e^{e^i} = e^{\cos 1}\sin(\sin 1)$ – achille hui Jul 17 '14 at 16:11
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If you have a new question, you should click on "ask question" and ask a new one; not edit the previous one. – Najib Idrissi Jul 18 '14 at 08:26
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@NajibIdrissi This isn't a new question..i know how this is done..i shared it because the answer is so elegant.. – user1001001 Jul 18 '14 at 08:29
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Well then, if this is an answer, you should post it as an answer to your question, and probably add more details... – Najib Idrissi Jul 18 '14 at 08:31
2 Answers
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Use the identity $\cos(n)+i\sin(n)=(\cos(1)+i\sin(1))^n$: $$\begin{aligned} \sum_{n=1}^\infty\frac{\sin(n)}{n!}&=\Im\sum_{n=1}^\infty\frac{\cos(n)+i\sin(n)}{n!}\\ &=\Im\sum_{n=1}^\infty\frac{(\cos(1)+i\sin(1))^n}{n!}\\ &=\Im e^{\cos(1)+i\sin(1)}\\ &=e^{\cos(1)}\sin(\sin(1)). \end{aligned}$$
Joffysloffy
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$$\sum_{n=1}^\infty \frac{\sin(n)}{n!} = \Im\sum\frac{e^{in}}{n!}=\Im e^{e^i}=\Im e^{\cos(1)}e^{i\sin(1)}=e^{\cos(1)}\sin(\sin(1)).$$
pshmath0
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1The symbols are different. Perhaps slightly more succinct. I prefer these steps myself. – pshmath0 Jul 21 '16 at 08:23