${\tan }^{-1} \left(\frac{1}{3}\right)+{\tan }^{-1} \left(\frac{2}{9}\right)+{\tan }^{-1} \left(\frac{4}{33}\right)+...+\mathrm{\infty }=$

The given expression is ${\tan }^{-1} \left(\frac{1}{3}\right)+{\tan }^{-1} \left(\frac{2}{9}\right)+{\tan }^{-1} \left(\frac{4}{33}\right)+...+\mathrm{\infty }$

This can be written as 

${\tan }^{-1}\left(\frac{2^1-2^0}{1+2^1·2^0}\right)+{\tan }^{-1}\left(\frac{2^2-2^1}{1+2^2·2^1}\right)+{\tan }^{-1}\left(\frac{2^3-2^2}{1+2^3·2^2}\right)+...\infty$

Use the formula ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)$

Hence, the given expression becomes 

${\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(4\right)-{\tan }^{-1}\left(2\right)+...+{\tan }^{-1}\left(\infty \right)$

Cancel out the terms and substitute the values 

$\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}=\boxed{\frac{\mathbf{\pi }}{\mathbf{4}}}$