If $S_n = \frac{n(n + 1)(n + 2)}{6}$,then $\sum_{n = 1}^\infty \frac{1}{t_n} = $

The sum of the series $\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n^2 - 1} + \sqrt{n^2}}$ equals

$\lim _{n \rightarrow \infty} \sum_{r=1}^n \tan ^{-1}\left(\frac{2 r}{r^4+r^2+2}\right) = $

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 $\sum\limits_{r = 1}^{10} {({r^2} + 1) \times r!}$ is equal to

If the sum of the first $10$ terms of the series $\frac{4(1)}{1+4(1)^4}+\frac{4(2)}{1+4(2)^4}+\frac{4(3)}{1+4(3)^4}+\ldots$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m+n$ is equal to . . . . . . .