4. ให้ $ \displaystyle{ f(x) = \frac{2}{2-x} = \frac{1}{1-\frac{x}{2}} = \sum_{n=0}^{\infty} (\frac{x}{2})^n, \ 0\leq x \leq 1 }$
จะได้ว่า
$\displaystyle{ 2 = f'(1) = \sum_{n=1}^{\infty} \frac{n}{2^n} }$
$\displaystyle{ 2\ln{2} - 1 = \int_{0}^{1} f(x) dx - 1 = \sum_{n=1}^{\infty} \frac{1}{2^n(n+1)} }$
ดังนั้น
$\displaystyle{ \sum_{n=1}^{\infty} \frac{n^2}{2^n(n+1)} = \sum_{n=1}^{\infty} \frac{n}{2^n} - \sum_{n=1}^{\infty} \frac{1}{2^n} + \sum_{n=1}^{\infty} \frac{1}{2^n(n+1)} = 2 - 1 + 2\ln{2} - 1 = 2\ln{2} }$
