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By the definition of the definite integral, the value of
$$
\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2-1}}+\frac{1}{\sqrt{n^2-2^2}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)
$$
is equal to
Options:
$$
\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2-1}}+\frac{1}{\sqrt{n^2-2^2}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)
$$
is equal to
Solution:
1277 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{2}$
Given that,
$\begin{aligned}
& \lim _{n \rightarrow \infty}\left\{\frac{1}{\sqrt{n^2-1^2}}+\frac{1}{\sqrt{n^2-2^2}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right\} \\
& \lim _{n \rightarrow \infty}\left\{\frac{1}{n \sqrt{1-\left(\frac{1}{n}\right)^2}}+\frac{1}{n \sqrt{1-\left(\frac{2}{n}\right)^2}}+\ldots+\frac{1}{n \sqrt{1-\left(\frac{n-1}{n}\right)^2}}\right\} \\
& \lim _{n \rightarrow \infty}\left\{\frac{1}{\sqrt{1-\left(\frac{1}{n}\right)^2}}+\frac{1}{\sqrt{1-\left(\frac{2}{n}\right)^2}}+\ldots+\frac{1}{\sqrt{1-\left(\frac{n-1}{n}\right)^2}}\right\} \\
& \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n-1} \frac{1}{n \sqrt{1-\left(\frac{1}{n}\right)^2}=\int_0^1 \frac{d x}{\sqrt{1-x^2}}}=\left[\sin ^{-1} x\right]_0^1 \\
& =\sin ^{-1} 1-\sin ^{-1} 0=\frac{\pi}{2}
\end{aligned}$
$\begin{aligned}
& \lim _{n \rightarrow \infty}\left\{\frac{1}{\sqrt{n^2-1^2}}+\frac{1}{\sqrt{n^2-2^2}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right\} \\
& \lim _{n \rightarrow \infty}\left\{\frac{1}{n \sqrt{1-\left(\frac{1}{n}\right)^2}}+\frac{1}{n \sqrt{1-\left(\frac{2}{n}\right)^2}}+\ldots+\frac{1}{n \sqrt{1-\left(\frac{n-1}{n}\right)^2}}\right\} \\
& \lim _{n \rightarrow \infty}\left\{\frac{1}{\sqrt{1-\left(\frac{1}{n}\right)^2}}+\frac{1}{\sqrt{1-\left(\frac{2}{n}\right)^2}}+\ldots+\frac{1}{\sqrt{1-\left(\frac{n-1}{n}\right)^2}}\right\} \\
& \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n-1} \frac{1}{n \sqrt{1-\left(\frac{1}{n}\right)^2}=\int_0^1 \frac{d x}{\sqrt{1-x^2}}}=\left[\sin ^{-1} x\right]_0^1 \\
& =\sin ^{-1} 1-\sin ^{-1} 0=\frac{\pi}{2}
\end{aligned}$
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