Why $\sum_{n=1}^\infty \frac{1}{2^{n}\ n} =\log (2)$?

Could someone explain me this equation? $$\sum_{n=1}^\infty \frac{1}{2^{n}\ n} =\log (2)$$ I am to calculate the sum of the expression on the left side. I was looking for mathematical formulas, any templates, but nothing was found. Have you got any idea?

• Hint: think about how you might evaluate $\sum_{n=1}^{\infty} \frac{1}{n2^n} x^n = \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{x}{2}\right)^n$ – Antonio Vargas Jul 26 '14 at 20:38
• Consider the Taylor series for $\log(1-x)$. – lemon Jul 26 '14 at 20:42
• Another method: Integrate both sides of sum of x^n/2^n = x/2/[1-x/2], manipulate terms a little and then substitute 1 for x. – berkeleychocolate Jul 26 '14 at 20:44
• @AntonioVargas do you have another hint? I'm interested in how you proceed on from there – DanZimm Jul 26 '14 at 20:52
• @DanZimm, See MathFacts' answer for more :) – Antonio Vargas Jul 26 '14 at 21:02

Since: $$\frac{x^{n+1}}{n+1}=\int_{0}^{x} y^n\,dy,$$ you have: $$S = \sum_{n=1}^{+\infty}\int_{0}^{1/2} y^{n-1}\,dy = \int_{0}^{1/2}\sum_{n=0}^{+\infty}y^n\,dy = \int_{0}^{1/2}\frac{dy}{1-y}=\left.-\log(1-y)\right|_{0}^{1/2}=\log 2.$$
• @oliveeuler no, we have $\sum_{n=1}^\infty \frac{1}{n2^n} = \sum_{n = 1}^\infty \int_0^{1 / 2} y^{n-1} dy$ – DanZimm Jul 26 '14 at 20:51
• @DanZimm I'd missed that the $y^{n}$ changed to $y^{n-1}$; my mistake. – Jam Jul 26 '14 at 20:53
Let $$f(x) = \sum_{n=1}^{\infty}\frac{x^n}{n2^n} \quad \Rightarrow \quad f^{\prime}(x) = \sum_{n=1}^{\infty}\frac{x^{n-1}}{2^n} = \frac{1}{x}\sum_{n=1}^{\infty}(x/2)^n = \frac{x/2}{x(1 - x/2)} = \frac{1}{2 - x}$$ Thus, $$f(x) = -\ln (2 - x) + C$$ For $x = 0$, $C = \ln 2$ and $f(x) = \ln \frac{2}{2 - x}$.
For $x = 1$, we have $$\sum_{n=1}^{\infty}\frac{1}{n2^n} = f(1) = \ln 2$$