# How to calculate $\sum_{k=1}^{k=n}\frac{\sin(kx)}{\sin^{k}(x)}$? (second question)

I asked yesterday for a hint on how to calculate $$1+\sum_{k=1}^{k=n}\frac{\sin(kx)}{\sin^{k}(x)}$$

I worked on this problem for another couple of hours and now I am stuck again, I would greatly appriciate another hint/help on how to continue.

This is what I done with the hint I got:

$$\sum_{k=1}^{k=n}\frac{\sin(kx)}{\sin^{k}(x)}=\frac{1}{2i}\sum_{k=1}^{k=n}((\frac{e^{xi}}{\sin(x)})^{k}-(\frac{e^{-xi}}{\sin(x)})^{k})=\frac{1}{2i}(\underbrace{\sum_{k=0}^{k=n}(\frac{e^{xi}}{\sin(x)})^{k}}_{S_{1}}-\underbrace{\sum_{k=0}^{k=n}(\frac{e^{-xi}}{\sin(x)}}_{S_{2}})^{k})$$

So I tried to calculate both sums and divide by $2i$:

$$S_{1}=\sum_{k=0}^{k=n}(\frac{e^{xi}}{\sin(x)})^{k}=1\cdot\frac{(\frac{e^{xi}}{\sin(x)})^{n+1}-1}{\frac{e^{xi}}{\sin(x)}-1}=\frac{\frac{e^{ix(n+1)}-\sin^{n+1}(x)}{\sin^{n+1}(x)}}{\frac{e^{xi}-\sin(x)}{\sin(x)}}=\frac{e^{ix(n+1)}-\sin^{n+1}(x)}{\sin^{n+1}(x)}\cdot\frac{\sin(x)}{e^{xi}-\sin(x)}=\frac{e^{ix(n+1)}-\sin^{n+1}(x)}{\sin^{n}(x)(e^{xi}-\sin(x))}$$

and similirly:

$$S_{2}=\sum_{k=0}^{k=n}(\frac{e^{-xi}}{\sin(x)})^{k}=1\cdot\frac{(\frac{e^{-xi}}{\sin(x)})^{n+1}-1}{\frac{e^{-xi}}{\sin(x)}-1}=\frac{\frac{e^{-ix(n+1)}-\sin^{n+1}(x)}{\sin^{n+1}(x)}}{\frac{e^{-xi}-\sin(x)}{\sin(x)}}=\frac{e^{-ix(n+1)}-\sin^{n+1}(x)}{\sin^{n+1}(x)}\cdot\frac{\sin(x)}{e^{-xi}-\sin(x)}=\frac{e^{-ix(n+1)}-\sin^{n+1}(x)}{\sin^{n}(x)(e^{-xi}-\sin(x))}$$

I then tooked the difference and arranged so I can divide by $2i$:

$$S_{1}-S_{2}=\frac{e^{ix(n+1)}-\sin^{n+1}(x)}{\sin^{n}(x)(e^{xi}-\sin(x))}-\frac{e^{-ix(n+1)}-\sin^{n+1}(x)}{\sin^{n}(x)(e^{-xi}-\sin(x))}=\frac{(e^{ix(n+1)}-\sin^{n+1}(x))(e^{-xi}-\sin(x))-((e^{xi}-\sin(x))(e^{-ix(n+1)}-\sin^{n+1}(x))}{\sin^{n}(x)(e^{-xi}-\sin(x))(e^{xi}-\sin(x))}$$

$$=\frac{e^{ixn}-\sin(x)e^{ix(n+1)}-\sin^{n+1}(x)e^{-xi}+\sin^{n+2}(x)-e^{-ixn}+\sin^{n+1}(x)e^{xi}+\sin(x)e^{-ix(n+1)}-\sin^{n+2}(x)}{\sin^{n}(x)(e^{-xi}-\sin(x))(e^{xi}-\sin(x))}$$

$$=\frac{e^{ixn}-e^{-ixn}-\sin(x)(e^{ix(n+1)}-e^{-ix(n+1)})+\sin^{n+1}(x)(e^{xi}-e^{-xi})}{\sin^{n}(x)(e^{-xi}-\sin(x))(e^{xi}-\sin(x))}$$

and now I divided by $2i$:

$$\implies\frac{s_{1}-s_{2}}{2i}=\frac{\sin(nx)-\sin(x)\sin(x(n+1))+\sin^{n+1}(x)\sin(x)}{\sin^{n}(x)(1-\sin(x)(e^{xi}+e^{-xi})+\sin^{2}(x))}$$

This is where I am stuck, I have high powers of $\sin(x)$ and also $\sin(nx)$ and $\sin(x(n+1))$ that I do not know what to do with, I keep in mind that since I have to add $1$ to this sum its probably meant to make the expression into something nicer, but if I add $1$ now I will do the oppisate

Any help or hint is appriciated!

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I don't exactly know what you are looking for, but clearly if you have to compute the given sum, say for $n=1000$, it will be easier to compute it with the second expression than with the first one. – Mercy King Nov 1 '12 at 19:10

It does not seem like you can get a much nicer expression. By the way, it might be easier to work it out by writing $\sin(kx) = \text{Im }e^{ikx}$. Then, since $\sin^k(x)$ is real, you can write
$$1 + \sum_{k=1}^{n} \frac{\sin(kx)}{\sin^k(x)} = 1 + \text{Im}\sum_{k=0}^n{\underbrace{\left(\frac{e^{ix}}{\sin(x)}\right)}_{f(x)}}^k = 1 + \text{Im} \frac{f(x)^{n+1} - 1}{f(x)-1}$$