<s>[INST] Evaluate $\sum_{n=1}^{\infty}{\frac{1}{n 2^{n-1}}}$ I need to find $$S = \sum_{n=1}^{\infty}{\frac{1}{n 2^{n-1}}}$$
Attempt:
$$f'(x) = \sum_{n=1}^{\infty}\frac{x^n}{2^n} = \frac{x}{2-x}$$
Which is just evaluating geometric series
$$f(x) = \sum_{n=1}^{\infty}\frac{x^{n+1}}{(n+1)2^{n}}$$
Now, by finding antiderivative of $\frac{x}{2-x}$
$$f(x) = -x-2\ln(x-2)$$
Finding sum should be just $$S = 1 + f(1)$$
But $f(x)$ is undefined as a real-valued function for $x \leq 2$
 [/INST] $$S=2\sum_{n=1}^{\infty} \frac{2^{-n}}{n}=2\sum_{k=1}^{\infty}~ 2^{-n} \int_{0}^{1} x^{n-1} ~dx=2 \int_{0}^{1}\sum_{n=1}^{\infty} \frac{dx}{x}\left(\frac{x}{2}\right)^n =\int_{0}^{1}\frac{2}{x} \frac{x/2}{1-x/2} ~dx=2 \ln 2. $$
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