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  4. The value of the displaystyle lim n arrow ∞[(1/√n2)+(1/√n2-1)+(1/√n2-22)+ ldots+(1/√n2-(n-1)2)] is

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Q. The value of the $\displaystyle\lim _{n \rightarrow \infty}\left[\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2-1}}+\frac{1}{\sqrt{n^2-2^2}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right]$ is

Integrals

Solution:

$ \underset{n \rightarrow \infty}{\text{Lim}} \left[\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2-1}}+\frac{1}{\sqrt{n^2-2^2}}+\ldots \ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right] $
$=\underset{n \rightarrow \infty}{\text{Lim}}\left[\frac{1}{\sqrt{n^2-0^2}}+\frac{1}{\sqrt{n^2-1^2}}+\ldots \ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right] $
$=\underset{n \rightarrow \infty}{\text{Lim}}\displaystyle\sum_{r=0}^{n-1} \frac{1}{\sqrt{n^2-r^2}}=\underset{n \rightarrow \infty}{\text{Lim}}\displaystyle\sum_{r=0}^{n-1} \frac{1}{n} \cdot \frac{1}{\sqrt{1-r^2 / n^2}} $
$=\int\limits_0^1 \frac{d x}{\sqrt{1-x^2}}=\left[\sin ^{-1} x\right]_0^1=\frac{\pi}{2} $