Show $\ln2 = \sum_\limits{n=1}^\infty\frac1{n2^n}$

Problem:

Show that $$\ln2 = \sum_\limits{n=1}^\infty\frac1{n2^n}.$$

My progress:

The problem before this one had me find the Taylor series for $\ln(1-x)$ which was $$-\sum\limits_{n=1}^\infty \frac{x^n}n$$

so I figured I'd use $x=-1$ and plug that into the Taylor series. However, there was a side note stating that the Taylor series I found is only valid for $x\in(-1, 1)$. And in any case, my calculation isn't going anywhere, since I end up with the series $1-\frac12+\frac13-\frac14\cdots$

Question 1: How can I use this to solve the problem stated initially, and in the title?

Question 2: I can see why $x$ is restricted to be less than 1, to prevent taking the log of zero or a negative. Why is it not valid for x greater than 1?

Solution 1:

Hint: $\ln 2 = - \ln (1/2)$. Can you use that and your result on the series for $\ln (1-x)$ to solve the problem?

Solution 2:

A different perspective: Your series is the Euler transform of $$ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots$$ and so it has an equal sum.

Solution 3:

Hint:

$$\frac1{1-x}=\sum_{n=0}^\infty x^n\;,\;\;|x|<1\implies-\log(1-x)=\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}=x+\sum_{n=2}^\infty\frac{x^n}n$$