Are you smarter than me? 5

Note first that $\dfrac{n+1}{n(n + 2)} = \dfrac{1}{2}*(\dfrac{1}{n} + \dfrac{1}{n+2})$ .

The sum can then be written as

$\displaystyle\sum_{n=1}^{\infty} (\dfrac{1}{2^{n+4}})*(\dfrac{1}{n} + \dfrac{1}{n+2}) =$

$(\dfrac{1}{16})*\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n*2^{n}} + (\dfrac{1}{4})*\sum_{n=1}^{\infty} \dfrac{1}{(n+2)*2^{n+2}} =$

$(\dfrac{1}{16})*\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n*2^{n}} + (\dfrac{1}{4})*(\sum_{n=1}^{\infty} \dfrac{1}{n*2^{n}} - \dfrac{1}{1*2^{1}} - \dfrac{1}{2*2^{2}}) =$

$(\dfrac{5}{16})*\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n*2^{n}} - \frac{5}{32} =$

( $\dfrac{5}{32})*(2\ln(2) - 1) = \boxed{0.0603584939.....}$ ,

where I used the fact that $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n*2^{n}} = \ln(2)$ , (equation $(19)$ here ).