The value of frac{1}{4} tan frac{pi}{8} + fra | vedclass

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(C) We use the identity $	an 	heta = \cot 	heta - 2 \cot 2	heta$. Let $S = \sum_{n=2}^{\infty} \frac{1}{2^n} 	an \frac{\pi}{2^{n+1}}$. Using the

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$\frac{2}{\pi} - \frac{1}{2}$

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(C) We use the identity $	an 	heta = \cot 	heta - 2 \cot 2	heta$. Let $S = \sum_{n=2}^{\infty} \frac{1}{2^n} 	an \frac{\pi}{2^{n+1}}$. Using the identity $	an 	heta = \cot 	heta - 2 \cot 2	heta$,we have $\frac{1}{2^n} 	an \frac{\pi}{2^{n+1}} = \frac{1}{2^n} \cot \frac{\pi}{2^{n+1}} - \frac{1}{2^{n-1}} \cot \frac{\pi}{2^n}$. This is a telescoping series. Summing from $n=2$ to $N$: $S_N = \sum_{n=2}^{N} \left( \frac{1}{2^n} \cot \frac{\pi}{2^{n+1}} - \frac{1}{2^{n-1}} \cot \frac{\pi}{2^n} ight)$. $S_N = \left( \frac{1}{2^N} \cot \frac{\pi}{2^{N+1}} - \frac{1}{2^1} \cot \frac{\pi}{2^2} ight)$. As $N 	o \infty$,$\frac{1}{2^N} \cot \frac{\pi}{2^{N+1}} = \frac{1}{2^N} \frac{\cos(\pi/2^{N+1})}{\sin(\pi/2^{N+1})} \approx \frac{1}{2^N} \frac{1}{\pi/2^{N+1}} = \frac{2}{\pi}$. Thus,$S = \frac{2}{\pi} - \frac{1}{2} \cot \frac{\pi}{4} = \frac{2}{\pi} - \frac{1}{2}$.

The value of $\frac{1}{4} \tan \frac{\pi}{8} + \frac{1}{8} \tan \frac{\pi}{16} + \frac{1}{16} \tan \frac{\pi}{32} + \dots \infty$ terms is equal to-

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