Sum $\sum_{n=1}^\infty\frac{n+2}{n(n+1)2^{n+1}}$

Question

I'm given the series $$\sum_{n=1}^\infty\frac{n+2}{n(n+1)2^{n+1}}$$

I've tried splitting up the fractions to use the telescoping effect, but no terms seem to cancel out. Can someone help me out?

Answer 1

The given expression can be written as $$\frac{n+2}{n(n+1)2^{n+1}} = \frac{1}{n2^n} - \frac{1}{(n+1)2^{n+1}}$$ which is the desired telescopic sum

Answer 2

Hint

Let $$\dfrac{n+2}{n(n+1)2^{n+1}}=f(n+1)-f(n)$$

where $f(m)=\dfrac{a_0+a_1m+a_2m^2+\cdots}{m2^m}$

$$\dfrac{n+2}{n(n+1)2^{n+1}}= \dfrac{n(a_0+a_1(n+1)+a_2(n+1)^2+\cdots)-2(n+1)(a_0+a_1n+a_2n^2+\cdots)}{n(n+1)2^{n+1}} $$

$$\implies n+2=n(a_0+a_1(n+1)+a_2(n+1)^2+\cdots)-2(n+1)(a_0+a_1n+a_2n^2+\cdots)$$

Compare the coefficients of $n^2$ and above to find $a_r=0\forall r\ge1$

$$\implies n+2=na_0-2(n+1)a_0=-2a_0-na_0$$

$$\implies a_0=-1$$