Question

The sum of the series $ta{n}^{-1}\frac{1}{2}+ta{n}^{-1}\frac{1}{8}+ta{n}^{-1}\frac{1}{18}+ta{n}^{-1}\frac{1}{32}+\cdots +\infty$ is

• A.
  
• B.
  
• C.
  
• D. None of these

Answer 1

The correct option is B

$ta{n}^{-1}\left(\frac{1}{2r^2}\right)=ta{n}^{-1}\left(\frac{2}{4r^2}\right)$
$=ta{n}^{-1}\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}$
$=ta{n}^{-1}(2r+1)-ta{n}^{-1}(2r-1)$
Thus,
${\sum }_{r=1}^{n}ta{n}^{-1}\left(\frac{1}{2r^2}\right)={\sum }_{r=1}^{n}ta{n}^{-1}(2r+1)-ta{n}^{-1}(2r-1)$
$=ta{n}^{-1}(2n+1)-ta{n}^{-1}(-1)$
$=ta{n}^{-1}(2n+1)-\frac{\pi }{4}$
$\underset{n\to \infty }{lim}ta{n}^{-1}(2n+1)-\frac{\pi }{4}$
$=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$