lim _{n rightarrow infty} n^4left[frac{1}{n^5 | vedclass

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Question

(C) આપેલ લક્ષ: $\lim _{n \rightarrow \infty} n^4 \sum_{r=0}^n \frac{1}{\left(n^2+r^2\right)^{\frac{5}{2}}}$$= \lim _{n \rightarrow \infty} \frac{1}{n}

Vedclass · Accepted Answer

$\frac{5}{6 \sqrt{2}}$

Vedclass · Answer

(C) આપેલ લક્ષ: $\lim _{n ightarrow \infty} n^4 \sum_{r=0}^n \frac{1}{\left(n^2+r^2ight)^{\frac{5}{2}}}$$= \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=0}^n \frac{n^5}{\left(n^2+r^2ight)^{\frac{5}{2}}}$$= \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=0}^n \frac{1}{\left(1+\left(\frac{r}{n}ight)^2ight)^{\frac{5}{2}}}$$= \int_0^1 \frac{1}{\left(1+x^2ight)^{\frac{5}{2}}} dx$ધારો કે $x = 	an 	heta$,તો $dx = \sec^2 	heta d	heta$. જ્યારે $x=0, 	heta=0$ અને જ્યારે $x=1, 	heta=\frac{\pi}{4}$.$I = \int_0^{\frac{\pi}{4}} \frac{\sec^2 	heta d	heta}{(\sec^2 	heta)^{\frac{5}{2}}} = \int_0^{\frac{\pi}{4}} \frac{\sec^2 	heta}{\sec^5 	heta} d	heta = \int_0^{\frac{\pi}{4}} \cos^3 	heta d	heta$$I = \int_0^{\frac{\pi}{4}} \cos 	heta (1 - \sin^2 	heta) d	heta$ધારો કે $\sin 	heta = t$,તો $\cos 	heta d	heta = dt$. જ્યારે $	heta=0, t=0$ અને જ્યારે $	heta=\frac{\pi}{4}, t=\frac{1}{\sqrt{2}}$.$I = \int_0^{\frac{1}{\sqrt{2}}} (1 - t^2) dt = \left[ t - \frac{t^3}{3} ight]_0^{\frac{1}{\sqrt{2}}}$$I = \frac{1}{\sqrt{2}} - \frac{1}{3(2\sqrt{2})} = \frac{1}{\sqrt{2}} - \frac{1}{6\sqrt{2}} = \frac{6-1}{6\sqrt{2}} = \frac{5}{6\sqrt{2}}$

$\lim _{n \rightarrow \infty} n^4\left[\frac{1}{n^5}+\frac{1}{\left(n^2+1\right)^{\frac{5}{2}}}+\frac{1}{\left(n^2+4\right)^{\frac{5}{2}}}+\frac{1}{\left(n^2+9\right)^{\frac{5}{2}}}+\ldots+\right]=$

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