Problem 4

Prove the following equality for any natural number

n

and any real number

x

for which no denominator vanishes (that is,

x\ne \frac{k\pi}{2^t}

for

t=0,1,\dots,n

and integer

k

\frac{1}{\sin 2x}+\frac{1}{\sin 4x}+\frac{1}{\sin 8x}+\cdots+\frac{1}{\sin 2^nx} = \cot x - \cot 2^nx.

Step 3 of 5: Telescope the sum

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In plain words

This is the standard telescoping pattern: a sum of consecutive differences collapses to (first value) minus (last value).

\sum_{k=1}^{n}\frac{1}{\sin 2^kx} = \sum_{k=1}^{n}\left[\cot(2^{k-1}x)-\cot(2^kx)\right] = \cot x - \cot(2^nx)

Detailed analysis

Adding the $n$ rewritten terms from step 2, every intermediate value $\cot(2^{k}x)$ for $1\le k\le n-1$ appears once with a $+$ sign and once with a $-$ sign and cancels, leaving only the first term $\cot x$ and the last term $-\cot(2^nx)$ .