Sum the following series:

Question:

Sum the following series:

$\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{2}{9}+\tan ^{-1} \frac{4}{33}+\ldots+\tan ^{-1} \frac{2^{n-1}}{1+2^{2 n-1}}$

Solution:

$\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{2}{9}+\tan ^{-1} \frac{4}{33}+\ldots+\tan ^{-1} \frac{2^{n-1}}{1+2^{2 n-1}}$

$\Rightarrow \tan ^{-1}\left(\frac{2-1}{1+2 \times 1}\right)+\tan ^{-1}\left(\frac{4-2}{1+4 \times 2}\right)+\tan ^{-1}\left(\frac{8-4}{1+8 \times 4}\right)+\ldots+\tan ^{-1}\left(\frac{2^{n}-2^{n-1}}{1+2^{n} \cdot 2^{n-1}}\right)$

$\Rightarrow\left(\tan ^{-1} 2-\tan ^{-1} 1\right)+\left(\tan ^{-1} 4-\tan ^{-1} 2\right)+\left(\tan ^{-1} 8-\tan ^{-1} 4\right)+\ldots+\left(\tan ^{-1} 2^{n-1}-\tan ^{-1} 2^{n-2}\right)+\left(\tan ^{-1} 2^{n}-\tan ^{-1} 2^{n-1}\right)$

$\Rightarrow \tan ^{-1} 2^{n}-\tan ^{-1} 1$

$\Rightarrow \tan ^{-1} 2^{n}-\frac{\pi}{4}$

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