QJEEAdv08Ma8ae2: JEE Advanced 2008 Mathematics - Class 11 JEEAdv08Ma8ae2 Solution: Let and for Then,

{S_n} < \mathop {\lim }\limits_{x \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{1 \over n}{1 \over {1 + {k \over n} + \left( {{k \over {{n^2}}}} \right)}}}

$= \int\limits_0^1 {{{dx} \over {1 + x + {x^2}}} = {\pi \over {3\sqrt 3 }}}$

As

h\sum\limits_{k = 0}^n {f(kh) > \int\limits_0^1 {f(x)dx > h} }

$\sum\limits_{k = 1}^n {f(kh)}$

So, {T_n} > {\pi \over {3\sqrt 3 }}