If $t_n = \frac{1}{4}(n+2)(n+3)$ for $n = 1, 2, 3, \ldots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}}$ is equal to

The expression $\frac{2^2+1}{2^2-1}+\frac{3^2+1}{3^2-1}+\frac{4^2+1}{4^2-1}+\ldots+\frac{(2011)^2+1}{(2011)^2-1}$ lies in the interval

$\frac{1}{1 \times 5} + \frac{1}{5 \times 9} + \frac{1}{9 \times 13} + \ldots$ up to $n$ terms $=$

If $S$ is the sum of the first $10$ terms of the series $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\tan ^{-1}\left(\frac{1}{21}\right)+\ldots$ then $\tan ( S )$ is equal to

The sum to $10$ terms of the series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\ldots$ is:

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