Question

Find the sum:
\[1+1+\frac{\sin2x}{\sin^2x}+\frac{\sin3x}{\sin^3x}+\cdots+\frac{\sin kx}{\sin^kx}\]

Answer 1

Hint: DeMoivre's theorem is your friend

Answer 2

@eliassaab

Answer 3

*

Answer 4

i wonder what summation this will have

Answer 5

This series does not seem to be even convergent for every x

Answer 6

what did you do to prove it's divergent?

Answer 7

1+1+sum (k=0 to oo)(sin(nx)/sin^n(x))

Answer 8

All the terms containing sin are infinite at x =0, or I might be missing something

Answer 9

how is DeMoiver will help

Answer 10

true i appears to me the limit is infinity for those terms?

Answer 11

it*

Answer 12

The original problem might have been a finite series up to \(k\) terms, I might be remembering it incorrectly.

Answer 13

Here's one approach. Let
\[S_{\text{sine}}=\sum_{n=1}^k\frac{\sin nx}{\sin^nx}=\frac{\sin x}{\sin x}+\frac{\sin2x}{\sin^2x}+\cdots+\frac{\sin kx}{\sin^kx}\]
and
\[S_{\text{cosine}}=\sum_{n=1}^k \frac{\cos nx}{\sin^nx}=\frac{\cos x}{\sin x}+\frac{\cos2x}{\sin^2x}+\cdots+\frac{\cos kx}{\sin^kx}\]
and so
\[\begin{align*}
S_{\text{cosine}}+iS_{\text{sine}}&=\frac{\cos x+i\sin x}{\sin x}+\frac{\cos2x+i\sin2x}{\sin^2x}+\cdots+\frac{\cos kx+i\sin kx}{\sin^kx}\\
&=\sum_{n=1}^k\frac{e^{inx}}{\sin^nx}
\end{align*}\]
(Now that I see it worked out again, I believe it was a finite series after all. Edited.) From here it's a matter of isolating the sine series.