mathop {lim }limits_{n to infty } ,sumlimits_ | vedclass

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(D) આપેલ લક્ષ $\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 0}^n \frac{n}{(2r + n)^2}$ છે.આપણે સરવાળાને આ રીતે લખી શકીએ: $\mathop {\lim }\l

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$\frac{1}{3}$

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(D) આપેલ લક્ષ $\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 0}^n \frac{n}{(2r + n)^2}$ છે.આપણે સરવાળાને આ રીતે લખી શકીએ: $\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 0}^n \frac{n}{n^2 (\frac{2r}{n} + 1)^2} = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r = 0}^n \frac{1}{(\frac{2r}{n} + 1)^2}$.નિશ્ચિત સંકલનની વ્યાખ્યા મુજબ,$\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r=0}^n f(\frac{r}{n}) = \int_0^1 f(x) dx$.અહીં,$x = \frac{r}{n}$ લેતા,સંકલન $\int_0^1 \frac{1}{(2x + 1)^2} dx$ બને છે.સંકલનનું મૂલ્ય: $\int_0^1 (2x + 1)^{-2} dx = [\frac{(2x + 1)^{-1}}{-1 	imes 2}]_0^1 = [-\frac{1}{2(2x + 1)}]_0^1$.સીમાઓ મૂકતા: $(-\frac{1}{2(3)}) - (-\frac{1}{2(1)}) = -\frac{1}{6} + \frac{1}{2} = \frac{-1 + 3}{6} = \frac{2}{6} = \frac{1}{3}$.

$\mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 0}^n {\frac{n}{{{{\left( {2r + n} \right)}^2}}}} $ ની કિંમત શોધો.

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