lim n → ∞ ( n n 4 + 1 + n n 4 + 16 + … n n 4 + n 4 ) − lim n → ∞ ( 2 n ( n 2 + 1 ) ( n 4 + 1 ) ) + 8 n ( n 2 + 4 ) n 4 + 1 + ⋯ 2 n ⋅ n 2 ( n 2 + n 2 ) n 4 + n 4 = lim n → ∞ ∑ r = 1 n 1 n 1 + r 4 n 4 − lim n → ∞ ∑ r = 1 n 1 n 2 ⋅ ( r / n ) 2 ( 1 + ( r n ) 2 ) 1 + r 4 n 4 = ∫ 0 1 d x 1 + x 4 − 2 ∫ 0 1 x 2 ( 1 + x 2 ) 1 + x 4 d x = ∫ 0 1 1 − x 2 ( 1 + x 2 ) 1 + x 4 d x \begin{aligned} & \lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}+\ldots \frac{n}{\sqrt{n^4+n^4}}\right) \\ & -\lim _{n \rightarrow \infty}\left(\frac{2 n}{\left(n^2+1\right)\left(\sqrt{n^4+1}\right)}\right)+ \\& \frac{8 n}{\left(n^2+4\right) \sqrt{n^4+1}}+\cdots \frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}} \\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n \sqrt{1+\frac{r^4}{n^4}}} \\& - \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n} \frac{2 \cdot(r / n)^2}{\left(1+\left(\frac{r}{n}\right)^2\right) \sqrt{1+\frac{r^4}{n^4}}} \\ & =\int_0^1 \frac{d x}{\sqrt{1+x^4}}-2 \int_0^1 \frac{x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x \\ & =\int_0^1 \frac{1-x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x \end{aligned} lim n → ∞ ( n 4 + 1 n + n 4 + 16 n + … n 4 + n 4 n ) − lim n → ∞ ( ( n 2 + 1 ) ( n 4 + 1 ) 2 n ) + ( n 2 + 4 ) n 4 + 1 8 n + ⋯ ( n 2 + n 2 ) n 4 + n 4 2 n ⋅ n 2 = lim n → ∞ r = 1 ∑ n n 1 + n 4 r 4 1 − lim n → ∞ r = 1 ∑ n n 1 ( 1 + ( n r ) 2 ) 1 + n 4 r 4 2 ⋅ ( r / n ) 2 = ∫ 0 1 1 + x 4 d x − 2 ∫ 0 1 ( 1 + x 2 ) 1 + x 4 x 2 d x = ∫ 0 1 ( 1 + x 2 ) 1 + x 4 1 − x 2 d x
∫ 0 1 ( 1 x 2 − 1 ) d x ( x + 1 x ) x 2 + 1 x 2 = ∫ 0 1 ( 1 x 2 − 1 ) d x ( x + 1 x ) ( x + 1 x ) 2 − 2 x + 1 x = t ⇒ ( 1 − 1 x 2 ) d x = d t − ∫ ∞ 2 d t t t 2 − 2 = ∫ 2 ∞ d t t t 2 − 2 = − 1 2 sin − 1 2 t ∣ 2 ∞ = − 1 2 ( 0 − π 4 ) = π 2 5 / 2 ∴ k = 2 5 / 2 ∴ k 2 = 32 \begin{aligned} & \int_0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{\left(x+\frac{1}{x}\right) \sqrt{x^2+\frac{1}{x^2}}} \\ & =\int_0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^2-2}} \\ & x+\frac{1}{x}=t \Rightarrow\left(1-\frac{1}{x^2}\right) 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}}=\left.\frac{-1}{\sqrt{2}} \sin ^{-1} \frac{\sqrt{2}}{t}\right|_2 ^{\infty} \\ & =\frac{-1}{\sqrt{2}}\left(0-\frac{\pi}{4}\right) \\ & =\frac{\pi}{2^{5 / 2}} \\ & \therefore k=2^{5 / 2} \\ & \therefore k^2=32 \end{aligned} ∫ 0 1 ( x + x 1 ) x 2 + x 2 1 ( x 2 1 − 1 ) d x = ∫ 0 1 ( x + x 1 ) ( x + x 1 ) 2 − 2 ( x 2 1 − 1 ) d x x + x 1 = t ⇒ ( 1 − x 2 1 ) d x = d t − ∫ ∞ 2 t t 2 − 2 d t = ∫ 2 ∞ t t 2 − 2 d t = 2 − 1 sin − 1 t 2 2 ∞ = 2 − 1 ( 0 − 4 π ) = 2 5 / 2 π ∴ k = 2 5 / 2 ∴ k 2 = 32
