I want to show
$$\log(2) = \sum_{k = 1}^{\infty} \frac{1}{k \cdot 2^{k}}.$$
So, I start with the geometric series:
$$\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n} $$
Then take the integral to get
$$\log(1 - x) = -\sum_{n = 1}^{\infty} \frac{x^{n}}{n}.$$
Then plug in $x = -1$ to get
$$\log(2) = -\sum_{n = 1}^{\infty} \frac{(-1)^{n}}{n} = \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}}{n},$$
but this isn't equal to the original equality. Why not? How can I do the problem?
Plug in $x=\dfrac12$ instead, and you will get the equality that you're after (since $\log\left(\frac12\right)=-\log(2)$).
Since the series are convergent for $|x| < 1$ then the "trick" is to use $x=1/2$. This leads to \begin{align} \frac{1}{1 - x} &= \sum_{n = 0}^{\infty} x^{n} \\ \log(1 - x) &= -\sum_{n = 1}^{\infty} \frac{x^{n}}{n} \end{align} becoming \begin{align} 2 &= \sum_{n=0}^{\infty} \frac{1}{2^{n}} \\ - \ln\left(1 - \frac{1}{2}\right) = \ln(2) &= \sum_{n=1}^{\infty} \frac{1}{2^{n} \, n} \end{align}
