The value of lim n k = 1 n n 3 ( n 2 + k 2 ) ( n 2 + 3 k 2 ) is: [2024]

Question

$The value of \lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n^{3}}{(n^{2} + k^{2}) (n^{2} + 3 k^{2})} is:$ [2024]

Smart Achievers · Accepted Answer

(4)

Let $L = \lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n^{3}}{(n^{2} + k^{2}) (n^{2} + 3 k^{2})}$

$= \lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n^{3}}{n^{4} (1 + \frac{k^{2}}{n^{2}}) (1 + \frac{3 k^{2}}{n^{2}})}$

$= \lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{(1 + \frac{k^{2}}{n^{2}}) (1 + \frac{3 k^{2}}{n^{2}})} = \int_{0}^{1} \frac{d x}{(1 + x^{2}) (1 + 3 x^{2})}$

$= \int_{0}^{1} [\frac{\frac{- 1}{2}}{(x^{2} + 1)} + \frac{\frac{3}{2}}{(3 x^{2} + 1)}] d x$

$= \frac{- 1}{2} \int_{0}^{1} \frac{1}{x^{2} + 1} d x + \frac{3}{2} \int_{0}^{1} \frac{1}{3 x^{2} + 1} d x$

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

$= \frac{- 1}{2} {[\tan^{- 1} x]}_{0}^{1} + \frac{\sqrt{3}}{2} {[\tan^{- 1} \sqrt{3} x]}_{0}^{1}$

$= \frac{- 1}{2} [\frac{π}{4} - 0] + \frac{\sqrt{3}}{2} [\frac{π}{3} - 0] = \frac{- π}{8} + \frac{π}{2 \sqrt{3}}$

$= \frac{π}{2 \sqrt{3}} - \frac{π}{8} = \frac{π}{2} (\frac{1}{\sqrt{3}} - \frac{1}{4}) = \frac{π (4 - \sqrt{3})}{8 \sqrt{3}}$

$= \frac{π (4 - \sqrt{3}) (4 + \sqrt{3})}{8 \sqrt{3} (4 + \sqrt{3})} = \frac{13 π}{8 (4 \sqrt{3} + 3)}$

Solution

Q.
$The value of \lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n^{3}}{(n^{2} + k^{2}) (n^{2} + 3 k^{2})} is:$ [2024]

1 $\frac{13 (2 \sqrt{3} - 3) π}{8}$

2 $\frac{π}{8 (2 \sqrt{3} + 3)}$

3 $\frac{(2 \sqrt{3} + 3) π}{24}$

4 $\frac{13 π}{8 (4 \sqrt{3} + 3)}$

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Solution

Q. The value of limn→∞∑k=1nn3(n2+k2)(n2+3k2) is: [2024]

1 13(23-3)π8

2 π8(23+3)

3 (23+3)π24