Calculate: $\frac{1}{4}\cdot(\sum_{n=1}^\infty (\frac{3}{4})^{n-1}\cdot\frac{1}{n})=\frac{2}{3} \ln2$

Question

I had a quiz in probability in which I solved a question and I got the following answer: $$\frac{1}{4}\cdot(\sum_{n=1}^\infty (\frac{3}{4})^{n-1}\cdot\frac{1}{n})$$

The right answer was $\frac{2}{3} \ln2$ which I know to be equivalent to the answer I got (WolframAlpha). How could one simplify/calculate this power series?

Answer 1

Approaching this from a probabilistic perspective, let $$f_{N}(n) = \dfrac{1}{4}\left(\dfrac{3}{4}\right)^{n-1}$$ for $n = 1, 2, \dots$. Then the sum is $$\sum_{n=1}^{\infty}\dfrac{1}{n}f_{N}(n)\text{,}$$ which is $\mathbb{E}\left[\dfrac{1}{N}\right]$, $N \sim \text{Geometric}$ with success probability $1/4$.

For general success probability $p \in (0, 1)$, use the answer at https://math.stackexchange.com/a/258090/81560.

Answer 2

If you start with the geometric series: $\Sigma_{n=1}^\infty x^{n-1}=\frac{1}{1-x}$ and then integrate both sides, you should be just about there.

Answer 3

Fill in details: for $\;|x|<1\;$ :

$$\frac1{1-x}=\sum_{n=0}^\infty x^n\implies-\log (1-x)=\sum_{n=1}^\infty\frac{x^n}n=x\sum_{n=1}^\infty\frac{x^{n-1}}n$$

Now substitute $\;x\;$ for some nice value and...etc.