lim _{n rightarrow infty} frac{1}{sqrt{n}}lef | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) આપેલ લક્ષ $L = \lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}} \sum_{r=1}^{n} \frac{1}{\sqrt{r}}$ છે.આપણે આ પદને $L = \lim _{n \rightarrow \infty}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) આપેલ લક્ષ $L = \lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}} \sum_{r=1}^{n} \frac{1}{\sqrt{r}}$ છે.આપણે આ પદને $L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{\sqrt{r/n}}$ તરીકે ફરીથી લખી શકીએ છીએ.નિશ્ચિત સંકલનની વ્યાખ્યા મુજબ,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$.અહીં,$f(x) = \frac{1}{\sqrt{x}}$ છે.તેથી,$L = \int_0^1 x^{-1/2} dx$.સંકલનનું મૂલ્ય શોધતા: $L = \left[ \frac{x^{1/2}}{1/2} \right]_0^1 = [2\sqrt{x}]_0^1 = 2(1) - 2(0) = 2$.

$\lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}}\left[1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}}\right]=$

Similar Questions

$\operatorname{Lim}_{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\ldots+\sin \frac{\pi}{2}\right]=$

$\lim _{n \rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}}{n \sqrt{n}}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{k}{{{n^2} + {k^2}}}} $ ની કિંમત શોધો.

$\lim _{n \rightarrow \infty} \frac{3}{n} \left\{ 4 + \left( 2 + \frac{1}{n} \right)^2 + \left( 2 + \frac{2}{n} \right)^2 + \dots + \left( 3 - \frac{1}{n} \right)^2 \right\}$ ની કિંમત શોધો.

$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2 j-1)+8 n}{(2 j-1)+4 n}$ નું મૂલ્ય શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module