lim _{n rightarrow infty}left[frac{n^{3 / 2}} | vedclass

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Question

(B) આપેલ લક્ષને સરવાળા તરીકે લખી શકાય છે:$\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \left[ \frac{n^{3/2}}{(n+2r)^{5/2}} - \frac{n^{1/2}}{(n+3r)^{3

Vedclass · Accepted Answer

$\frac{-1}{9 \sqrt{3}}$

Vedclass · Answer

(B) આપેલ લક્ષને સરવાળા તરીકે લખી શકાય છે:$\lim _{n ightarrow \infty} \sum_{r=0}^{n-1} \left[ \frac{n^{3/2}}{(n+2r)^{5/2}} - \frac{n^{1/2}}{(n+3r)^{3/2}} ight]$અંશ અને છેદને અનુક્રમે $n^{5/2}$ અને $n^{3/2}$ વડે ભાગતા:$\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} \left[ \frac{1}{(1+2r/n)^{5/2}} - \frac{1}{(1+3r/n)^{3/2}} ight]$આ રીમાન સરવાળો છે જે નિશ્ચિત સંકલનમાં રૂપાંતરિત થાય છે:$\int_0^1 \left( \frac{1}{(1+2x)^{5/2}} - \frac{1}{(1+3x)^{3/2}} ight) dx$પદવાર સંકલન કરતા:$= \left[ \frac{(1+2x)^{-3/2}}{-3/2 	imes 2} - \frac{(1+3x)^{-1/2}}{-1/2 	imes 3} ight]_0^1$$= \left[ -\frac{1}{3(1+2x)^{3/2}} + \frac{2}{3(1+3x)^{1/2}} ight]_0^1$$0$ અને $1$ સીમાઓ પર મૂલ્યાંકન કરતા:$= \left( -\frac{1}{3(3)^{3/2}} + \frac{2}{3(4)^{1/2}} ight) - \left( -\frac{1}{3(1)^{3/2}} + \frac{2}{3(1)^{1/2}} ight)$$= \left( -\frac{1}{9\sqrt{3}} + \frac{2}{6} ight) - \left( -\frac{1}{3} + \frac{2}{3} ight)$$= -\frac{1}{9\sqrt{3}} + \frac{1}{3} - \frac{1}{3} = -\frac{1}{9\sqrt{3}}$

$\lim _{n \rightarrow \infty}\left[\frac{n^{3 / 2}}{n^{5 / 2}}-\frac{n^{1 / 2}}{n^{3 / 2}}+\frac{n^{3 / 2}}{(n+2)^{5 / 2}}-\frac{n^{1 / 2}}{(n+3)^{3 / 2}}+\ldots+\frac{n^{3 / 2}}{(n+2(n-1))^{5 / 2}}-\frac{n^{1 / 2}}{(n+3(n-1))^{3 / 2}}\right]=$

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જો $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{r^4+n^4}=p$ હોય,તો $e^p=$

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