Question

If $ta{n}^{-1}\frac{2m}{m^4+m^2+2}=a_m,then{a}_1+a_2+a_3+.........to\infty =$

• A. $\frac{\pi }{4}$
• B. $-\frac{\pi }{4}$
• C. $\frac{3\pi }{4}$
• D. None

Answer 1

The correct option is A $\frac{\pi }{4}$

$ta{n}^{-1}\frac{2m}{2+m^2+m^4}=ta{n}^{-1}\frac{2m}{1+(m^4+m^2+1)}=ta{n}^{-1}\frac{(m^2+m+1)-(m^2-m+1)}{1+(m^2+m+1)(m^2-m+1)}=ta{n}^{-1}(m^2+m+1)-ta{n}^{-1}(m^2-m+1)$
Let $S_n={\sum }_{m=1}^{n}ta{n}^{-1}\frac{2m}{2+m^2+m^4}=(ta{n}^{-1}3=ta{n}^{-1}1)+(ta{n}^{-1}7-ta{n}^{-1}3)+(ta{n}^{-1}3-ta{n}^{-1}7)+......+ta{n}^{-1}(n^2+n+1)-ta{n}^{-1}(n^2-n+1)=ta{n}^{-1}(n^2+n+1)-ta{n}^{-1}1$
As $n\to \infty ,ta{n}^{-2}(n^2+n+1)\to \frac{\pi }{2}$
as $n^2+n+1\to \infty$
Hence $S_{\infty }=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$.