JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 13)
Let $${S_n} = \sum\limits_{k = 1}^n {{n \over {{n^2} + kn + {k^2}}}} $$ and $${T_n} = \sum\limits_{k = 0}^{n - 1} {{n \over {{n^2} + kn + {k^2}}}} $$ for $$n$$ $$=1, 2, 3, ............$$ Then,
$${S_n} < {\pi \over {3\sqrt 3 }}$$
$${S_n} > {\pi \over {3\sqrt 3 }}$$
$${T_n} < {\pi \over {3\sqrt 3 }}$$
$${T_n} > {\pi \over {3\sqrt 3 }}$$
Explanation
$${S_n} < \mathop {\lim }\limits_{x \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{1 \over n}{1 \over {1 + {k \over n} + \left( {{k \over {{n^2}}}} \right)}}} $$
$$ = \int\limits_0^1 {{{dx} \over {1 + x + {x^2}}} = {\pi \over {3\sqrt 3 }}} $$
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, $${T_n} > {\pi \over {3\sqrt 3 }}$$
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