Let $$ S_{n}=\sum_{k=1}^{n} \frac{n}{n^{2}+k n+k^{2}} \quad \text { and } \quad T_{n}=\sum_{k=0}^{n-1} \frac{n}{n^{2}+k n+k^{2}}, $$ for $n=1,2,3, \cdots$. Then,

(A) $S_{n} \frac{\pi}{3 \sqrt{3}}$

(B) $S_{n}>\frac{\pi}{3 \sqrt{3}}$

(C) $T_{n}<\frac{\pi}{3 \sqrt{3}}$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Definite Integrals

Let

S_{n} = k = 1 \sum n \frac{n}{n ^{2} + k n + k ^{2}} and T_{n} = k = 0 \sum n - 1 \frac{n}{n ^{2} + k n + k ^{2}},

for

n = 1, 2, 3, \dots

. Then,

Answer: (A), (D)

We can interpret these sums as Riemann sums.

For

S_{n}

S_{n} = k = 1 \sum n \frac{1}{n} \cdot \frac{1}{1 + k / n + ( k / n ) ^{2}}

n \to \infty

, this is a right Riemann sum for:

\int_{0}^{1} \frac{d x}{1 + x + x ^{2}}

To evaluate:

\int_{0}^{1} \frac{d x}{1 + x + x ^{2}} = \int_{0}^{1} \frac{d x}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}}

Let

u = x + \frac{1}{2}

= \int_{1/2}^{3/2} \frac{d u}{u ^{2} + 3/4} = \frac{2}{3} [tan^{- 1} (\frac{2 u}{3})]_{1/2}^{3/2} = \frac{2}{3} [tan^{- 1} (3) - tan^{- 1} (\frac{1}{3})]

= \frac{2}{3} [\frac{π}{3} - \frac{π}{6}] = \frac{2}{3} \cdot \frac{π}{6} = \frac{π}{3 3}

Since

S_{n}

is a right Riemann sum and the function is decreasing,

S_{n} < \int_{0}^{1} \frac{d x}{1 + x + x ^{2}} = \frac{π}{3 3}

.

Similarly,

T_{n}

is a left Riemann sum, so

T_{n} > \frac{π}{3 3}

.

Therefore, (A) and (D) are correct.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · One or more correct options