I need to solve this sum, but unfortunately I have no idea how to start.I would be grateful for every advice.
$$\sum _{n=1}^{\infty }{\left(q\right)^n\left(\sin(na)\right), |q|<1}$$
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Chinnapparaj R
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Many
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3Try the Euler formula $\mathrm e^{\mathrm i t} = \cos t + \mathrm i \sin t$. – xbh Aug 25 '19 at 09:49
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This type of question was asked before, did you search before asking? – rtybase Aug 25 '19 at 09:54
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Like here or here for example. – rtybase Aug 25 '19 at 10:01
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1Possible duplicate of Sum of $\sum\limits_{n=1}^\infty q^n \sin(nx)$ – Arnaud D. Aug 25 '19 at 14:12
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As a general tip, I recommend the use of Approach0 to find duplicates. For example, see what it gives in this case : https://approach0.xyz/search/?q=%24%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7Dq%5En%5Csin%5Cleft(na%5Cright)%24&p=1 – Arnaud D. Aug 25 '19 at 14:14
3 Answers
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It's the imaginary part of $$\sum_{n=0}^\infty q^n e^{ina}=\frac1{1-qe^{ia}}.$$
Angina Seng
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As $\;\sin nt=\operatorname{Im}\mathrm e^{int}$, we have $$\sum _{n=1}^{\infty }(q )^n \sin(na)=\operatorname{Im}\Bigl(\sum _{n=1}^{\infty }(q \mathrm e^{it})^n\Bigr).$$ Also, we have $$\sum _{n=1}^{\infty } z^n=z\sum _{n=0}^{\infty } z^n=\frac z{1-z},\quad |z|<1.$$
Bernard
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The complex method is probably the best.
But instead you can: prove by induction on $N$ that
$$
\sum _{n=1}^{N}{q}^{n}\sin \left( na \right) ={\frac {{q}^{N+2}\sin
\left( Na \right) -{q}^{N+1}\sin \left( \left( N+1 \right) a
\right) +q \sin \left( a \right)}{{q}^{2}-2q\cos \left( a \right)+
1}}
$$
and then do the limit easily:
$$
\sum _{n=1}^{\infty}{q}^{n}\sin \left( na \right) = {\frac {q \sin \left( a \right)}{{q}^{2}-2q\cos \left( a \right)+
1}}
$$
GEdgar
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