Test the convergence of the series 1/3.7+1/4.9+1/5.11+… OR Investigate convergence of the series 1/3.7+1/4.9+1/5.11+…
Solution Here
u_n=\frac{1}{(n+2)(2n+5)}
Let v_n=\frac{1}{n^2} now consider \frac{u_n}{v_n}=\frac{1}{(n+2)(2n+5)}n^2={\frac{n^2}{2n^2+9n+10}}
\Rightarrow \lim_{n\rightarrow \infty}\frac{u_n}{v_n}=\lim_{n\rightarrow \infty}\frac{n^2}{2n^2+9n+10}
\lim_{n\rightarrow \infty}\frac{1}{2+\frac{9}{n}+\frac{10}{n^2}}=\frac{1}{2}
(which is a finite and non zero number)
Hence by Limit form test , \sum_{n=0}^{\infty}u_n and \sum_{n=0}^{\infty}v_n behave similarly.
\sum_{n=0}^{\infty}v_n=\sum_{n=0}^{\infty}\frac{1}{n^2} converges (as p=2 > 1)
\therefore \sum_{n=0}^{\infty}u_n=\sum_{n=0}^{\infty}\frac{1}{(n+2)(2n+5)} also converges.
\frac{1}{3}+\frac{1.2}{3.5}+\frac{1.2.3}{3.5.7}+...\infty