Let lim n ( n n 4 + 1 - 2 n ( n 2 + 1 ) n 4 + 1 + n n 4 + 16 - 8 n ( n 2 + 4 ) n 4 + 16 + ? + n n 4 + n 4 - 2 n · n 2 ( n 2 + n 2 ) n 4 + n 4 ) be k , using only the principal values of the inverse trigonometric functions. Then k 2 is equal to ______

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Solution

Q.
Let $\lim_{n \to \infty} (\frac{n}{\sqrt{n^{4} + 1}} - \frac{2 n}{(n^{2} + 1) \sqrt{n^{4} + 1}} + \frac{n}{\sqrt{n^{4} + 16}} - \frac{8 n}{(n^{2} + 4) \sqrt{n^{4} + 16}} + \dots + \frac{n}{\sqrt{n^{4} + n^{4}}} - \frac{2 n \cdot n^{2}}{(n^{2} + n^{2}) \sqrt{n^{4} + n^{4}}})$ be $\frac{π}{k},$ using only the principal values of the inverse trigonometric functions. Then $k^{2}$ is equal to ______ . [2024]

Ans.
(32)

$\lim_{n \to \infty} (\frac{n}{\sqrt{n^{4} + 1}} + \frac{n}{\sqrt{n^{4} + 16}} + \dots \frac{n}{\sqrt{n^{4} + n^{4}}}) - \lim_{n \to \infty} (\frac{2 n}{(n^{2} + 1) (\sqrt{n^{4} + 1})} + \frac{8 n}{(n^{2} + 4) \sqrt{n^{4} + 16}} + \dots \frac{2 n \cdot n^{2}}{(n^{2} + n^{2}) \sqrt{n^{4} + n^{4}}})$

$= \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n \sqrt{1 + \frac{r^{4}}{n^{4}}}} - \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} \frac{2 \cdot {(\frac{r}{n})}^{2}}{(1 + {(\frac{r}{n})}^{2})} \frac{1}{\sqrt{1 + \frac{r^{4}}{n^{4}}}}$

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

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

$Put x + \frac{1}{x} = t, then (1 - \frac{1}{x^{2}}) d x = d t$

$= - \int_{\infty}^{2} \frac{d t}{t \sqrt{t^{2} - 2}} = \int_{2}^{\infty} \frac{d t}{t \sqrt{t^{2} - 2}}$

$Put t^{2} - 2 = u^{2} \Rightarrow 2 t d t = 2 u d u \Rightarrow d t = \frac{u}{t} d u$

$= \int_{\sqrt{2}}^{\infty} \frac{d u}{(u^{2} + 2)} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{u}{\sqrt{2}}$ $|_{\sqrt{2}}^{\infty}$

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

$\Rightarrow k = 2^{5 / 2} \Rightarrow k^{2} = 2^{5} = 32$

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