Answer
Convergent with sum $\dfrac{5}{12}$
Work Step by Step
We are given the series:
$\sum_{n=2}^{\infty}\dfrac{1}{n(n+2)}$.
A partial sum for m terms can be written:
$S_m=\sum_{n=2}^{m}\dfrac{1}{n(n+2)}=\sum_{n=2}^{m}\left(\dfrac{A}{n}+\dfrac{B}{n+2}\right)$
Determine $A,B$:
$A(n+2)+Bn=1$
$An+2A+Bn=1$
$(A+B)n+2A=1$
$2A=1\Rightarrow A=\dfrac{1}{2}$
$A+B=0\Rightarrow B=-A=-\dfrac{1}{2}$
Rewrite $S_m$:
$S_m=\sum_{n=2}^{m}\left(\dfrac{\dfrac{1}{2}}{n}+\dfrac{-\dfrac{1}{2}}{n+2}\right)=\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)$
$=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+.....+\dfrac{1}{m}-\dfrac{1}{m+2}\right)$
$=\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{m+1}-\dfrac{1}{m+2}\right)$
Determine at least 10 partial sums:
$S_2=\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{3}-\dfrac{1}{4}\right)= 0.625$
$S_3=\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{4}-\dfrac{1}{5}\right)\approx 0.19167$
$S_4=\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{5}-\dfrac{1}{6}\right)\approx 0.23333$
$S_5=\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{6}-\dfrac{1}{7}\right)\approx 0.26190$
$S_{10}=\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{11}-\dfrac{1}{12}\right)\approx 0.32955$
$S_{25}=\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{26}-\dfrac{1}{27}\right)\approx 0.37892$
$S_{50}=\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{51}-\dfrac{1}{52}\right)\approx 0.39725$
$S_{100}=\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{101}-\dfrac{1}{102}\right)\approx 0.40681$
$S_{500}=\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{501}-\dfrac{1}{502}\right)\approx 0.41467$
$S_{1000}=\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{1001}-\dfrac{1}{1002}\right)\approx 0.41567$
Graph both the sequence of terms (in blue color) and the sequence of partial sums (in red color).
When $m\rightarrow \infty$, $\dfrac{1}{m+1}\rightarrow 0$ and $\dfrac{1}{m+2}\rightarrow 0$, so $S_m\rightarrow \dfrac{1}{2}\left(\dfrac{5}{6}\right)=\dfrac{5}{12}$, therefore the series is convergent, its sum being $\dfrac{5}{12}$.
