Express this series as a telescoping series then evaluate. The image displays a mathematical series expressed as:\[\sum_{n=1}^{\infty} \frac{n+2}{n(n+1)2^{n+1}}\]This is an infinite series where the general term is given by the fraction \(\frac{n+2}{n(n+1)2^{n+1}}\). Here, the numerator is \(n+2\), and the denominator is the product of \(n\), \(n+1\), and \(2^{n+1}\). The series is summed from \(n=1\) to infinity.

Consider the series ∑n=1∞n+2nn+12n+1 First we will express the given series in the form of telescopic series ∑n=1∞n+2nn+12n+1=∑n=1∞n+2+n−nnn+12n+1Add and substract n from numenator=∑n=1∞2n+2−nnn+12n+1=∑n=1∞2n+2nn+12n+1−nnn+12n+1=∑n=1∞2n+1nn+12n×2−nnn+12n+1=∑n=1∞1n2n−1n+12n+1 Hence, ∑n=1∞n+2nn+12n+1=∑n=1∞1n2n−1n+12n+1 Which is required telescopic series formNow, we will find the sum of the given series by using sequence of partial some Let, an=1n2n−1n+12n+1 Substitute n=1,2,3,....... a1=11⋅21−12⋅22a2=12⋅22−13⋅23a3=13⋅23−14⋅24...an=1n2n−1n+1⋅2n+1Now, adding the corresponding term, we get a1+a2+a3+.....+an=12−1n+1⋅2n+1 Let Sn be the sequence of partial sum, which is defined as Sn=limn→∞a1+a2+a3+.....+an Sn=limn→∞a1+a2+a3+.....+an=limn→∞12−1n+1⋅2n+1=12−0=12 Hence, Sn=12 Therefore, ∑n=1∞n+2nn+12n+1=12