The sum of $(n + 1)$ terms of $\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ is

The value of the infinite product $\prod\limits_{n = 2}^\infty {\left( {1 - \frac{1}{{{n^2}}}} \right)}$ is

$\frac{1}{3 \cdot 6} + \frac{1}{6 \cdot 9} + \frac{1}{9 \cdot 12} + \dots$ to $9$ terms $=$

If ${T_n} = ({n^2} + 1)n!$ and ${S_n} = {T_1} + {T_2} + {T_3} + ...... + {T_n}$. Let $\frac{{{T_{10}}}}{{{S_{10}}}} = \frac{a}{b}$ where $a$ and $b$ are relatively prime natural numbers,then the value of $(b - a)$ is

The sum of $n$ terms of the series $\frac{1}{1 + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \dots$ is

The value of $\sum\limits_{n = 1}^\infty {\left( {{{\tan }^{ - 1}}\left( {\frac{n}{{n + 2}}} \right) - {{\tan }^{ - 1}}\left( {\frac{{n - 1}}{{n + 1}}} \right)} \right)} $ is equal to-