# How to show that $\sum_{i=1}^\infty\frac{1}{i\cdot2^i}=\ln2$? [closed]

I have no idea about showing that:

$$\sum_{i=1}^\infty\frac{1}{i\cdot2^i}=\ln2$$

And what about more general situation(replace 2 with a constant $$\alpha$$)?

Could anyone please give me a helping hand? Any help would be appreciated.

• You know that the dummy variable is $i$, but in the sum you typed $x$, right? Also maybe you should start at $i=1$. – Zacky Feb 10 '19 at 15:58
• As a hint, think about the power series of $\ln(1-x)$. – Zacky Feb 10 '19 at 16:02
Start with the geometric series. For $$|x|<1$$ it holds $$\sum_{k=0}^\infty x^k =\frac{1}{1-x}$$ Integration renders: $$\sum_{k=1}^\infty \frac{1}{k}x^{k} =-ln (1-x)$$ Setting $$x=1/2$$ renders your formula.
At first, start by using the power series of $$ln(1-x)$$. Then,
For a more general constant (say $$a$$), use $$1/a$$ instead of $$1/2$$. But this is true as long as $$|a| \geq 1$$
The basic question has been answered. A generalization is $$\sum_{k=1}^\infty\frac{1}{k\alpha^k}=\ln(\frac \alpha {\alpha-1})$$