Sum of product of AP, GP, HP. $\sum n \bigg( \frac{1}{2^n} \bigg) \bigg( \frac{1}{n+1} \bigg)$

Question

How to find the sum of the first $n$ terms, and the infinite sum, of the series

$$\frac{1}{4}+\frac{1}{6}+\frac{3}{32}+\frac{1}{20}+\frac{5}{192}+\dots + n \bigg( \frac{1}{2^n} \bigg) \bigg( \frac{1}{n+1} \bigg) + \dots \text{ ?}$$

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I know that the sum of arithmetic series is $S_n=\frac{n}{2} \big(2a_1+d(n-1) \big)$, where $a_1$ is the first term, and $d$ is the common difference. Also, I know that the sum of the geometric series is $S_n=a_1 \big(\frac{r^n-1}{r-1} \big)$, where $a_1$ is the first term, and $r$ is the common ratio. Furthermore, I know that numbers are in harmonic progression if their reciprocals are in arithmetic progression.

I took a look at Arithmetico–Geometric Sequence. Still, I do not know how to find the sum of the given expression. Each term in the given expression is a product of $3$ factors, the first factor in each term is AP, the second factor in each term is GP, and the third factor in each term is HP.

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Wolfram Alpha says that the infinite sum is $2\big(1-\log(2)\big)$.

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Your help would be appreciated. THANKS.

Answer 1

Hint: $n/(n+1)=1-1/(n+1)$. This breaks the sum into two. The first you know. The second is directly obtainable through the Taylor expansion of $\ln x$ at 1.