A) \[\frac{\pi }{2\sqrt{2}}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{8}\]
Correct Answer: C
Solution :
| We have, \[_{n\to \infty }^{\lim }\sum\limits_{r=0}^{n}{\left( \frac{1}{4r+1}-\frac{1}{4r+3} \right)}\]\[=_{n\to \infty }^{\lim }\sum\limits_{r=0}^{n}{\left( \int_{0}^{1}{{{x}^{4r}}-{{x}^{4r+2}}} \right)dx}\] |
| \[=\int_{0}^{1}{\left( 1+{{x}^{4}}+{{x}^{8}}+.... \right)-\left( {{x}^{2}}+{{x}^{6}}+{{x}^{10}}... \right)}dx\] |
| \[=\int_{0}^{1}{\left( \frac{1}{1-{{x}^{4}}}-\frac{{{x}^{2}}}{1-{{x}^{4}}} \right)dx}\] |
| \[=\int_{0}^{1}{\frac{1-{{x}^{2}}}{1-{{x}^{4}}}}dx=\int_{0}^{1}{\frac{dx}{1+{{x}^{2}}}}\] |
| \[=\left[ {{\tan }^{-1}}x \right]_{0}^{1}=\frac{\pi }{4}\] |
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