3

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.

Max
  • 1,039
ArchanJin
  • 73
  • 2
    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
  • 2
    As a hint, think about the power series of $\ln(1-x)$. – Zacky Feb 10 '19 at 16:02
  • 1
    Also, please search on approach0.xyz before asking, since often your question has been asked and answered before. – ViktorStein Feb 10 '19 at 16:40

3 Answers3

3

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.

Maksim
  • 1,706
3

At first, start by using the power series of $ln(1-x)$. Then,

enter image description here For a more general constant (say $a $), use $1/a $ instead of $1/2$. But this is true as long as $|a| \geq 1$

shsh23
  • 1,135
1

The basic question has been answered. A generalization is $$\sum_{k=1}^\infty\frac{1}{k\alpha^k}=\ln(\frac \alpha {\alpha-1})$$

J. W. Tanner
  • 60,406