Question

$\underset{n\to \infty }{lim}\frac{1}{\sqrt{n^2-1}}+\frac{1}{\sqrt{n^2-4}}+\frac{1}{\sqrt{n^2-9}}+.....\frac{1}{\sqrt{2n-1}}=$

• A. $\pi /4$
• B. $\log 2$
• C. $\pi /2$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

The correct option is C

$\pi /2$

$\underset{n\to \infty }{lim}\frac{1}{\sqrt{n^2-1}}+\frac{1}{\sqrt{n^2-4}}+\frac{1}{\sqrt{n^2-9}}+.....\frac{1}{\sqrt{2n-1}}=\underset{n\to \infty }{lim}\sum _{r=1}^{n-1}\frac{1}{\sqrt{n^2-r^2}}$

$=\underset{n\to \infty }{lim}\sum _{r=1}^{n-1}\frac{1/n}{\sqrt{1-(r/n{)}^2}}$

= $\underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi }{2}$