# How do I show that $\ln(2) = \sum \limits_{n=1}^\infty \frac{1}{n2^n}$?

Show that $$\ln(2) = \sum \limits_{n=1}^\infty \frac{1}{n2^n}$$

My work so far:

If we approximate $\ln(x)$ around $x = 1$, we get:

$\ln(x) = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + ...$

Substituting $x = 2$ then gives us:

$\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + ...$

No suprise there, we should always get alternating series for $\ln(x)$ when we are doing the taylor expansion. By using Euler transform which is shown at the middle of this page on natural logarithm one can obtain the wanted result, but how do you derive it? I need someone to actually show and explain in detail how one starts from the left side of the equation and ends up on the other.

• Well, your series is expanded in powers of $\frac{1}{2}$, and your $\ln x$ series is expanded in powers of $-(x-1)$ (noting that the signs of the coefficients alternate) - just choose a suitable $x$ to make these match up. – πr8 May 9 '16 at 9:55
• Hint: $\ln(\frac{x}{x-1})=-\ln(\frac{x-1}{x})=-\ln(1-\frac{1}{x})$ – MrYouMath May 9 '16 at 9:57
• – Martin Sleziak Aug 30 '16 at 23:13

Note that \begin{align} \frac{1}{1-x}&=1+x+x^2+\cdots\qquad(-1<x<1). \end{align} Thus \begin{align} -\ln(1-x)&=\int\frac{1}{1-x}{\rm d}x\\ &=x+\frac{1}{2}x^2+\frac{1}{3}x^3+\cdots\\ &=\sum_{n=1}^\infty\frac{1}{n}x^n. \end{align} By taking $x=\frac{1}{2}$ into the above equation, we have $$\sum_{n=1}^\infty\frac{1}{n2^n}=-\ln\left(\frac{1}{2}\right)=\ln(2).$$
\begin{align} \sum \limits_{n=1}^\infty \frac{1}{n2^n}&=\sum \limits_{n=1}^\infty \int_0^{1/2}x^{n-1}dx \\\\&=\int_0^{1/2}\sum \limits_{n=1}^\infty x^{n-1}\:dx \\\\&=\int_0^{1/2}\frac1{1-x}\:dx \\\\&=\left[-\ln(1-x)\right]_0^{1/2} \\\\&=\ln 2 \end{align}
Similarly, one gets $$\sum \limits_{n=1}^\infty \frac{t^n}{n}=-\ln(1-t),\quad |t|<1.$$