Find the sum to $n$ terms of the series $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots$

The value of $\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\frac{3}{{{r^2} - r + 9}}} \right)} $ is-

Let for $n = 1, 2, \ldots, 50$,$S_{n}$ be the sum of the infinite geometric progression whose first term is $n^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26} + \sum_{n=1}^{50} \left(S_{n} + \frac{2}{n+1} - n - 1\right)$ is equal to

If $t_{n} = \frac{1}{4}(n+2)(n+3)$,$n \in N$,then which one of the following is true?
Assertion $(A)$ : $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}} = \frac{2003}{3009}$
Reason $(R)$ : $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{n}} = \frac{4n}{3(n+3)}$

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

If ${t_n} = \frac{1}{4}(n + 2)(n + 3)$ for $n = 1, 2, 3, \dots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \dots + \frac{1}{t_{2003}} = $