# Thread: convergence problem

1. ## convergence problem

I'm trying to understand how to get to this equality step by step. I appreciate your help.

2. ## Re: convergence problem

What have you tried? Are you familiar with geometric progressions? What happens when you take the derivative term-by-term?

3. ## Re: convergence problem

Originally Posted by SlipEternal
What have you tried? Are you familiar with geometric progressions? What happens when you take the derivative term-by-term?
I meant geometric sums of the form:

$$\sum_{k=0}^\infty r^k = \dfrac{1}{1-r}, |r|<1$$

4. ## Re: convergence problem

Yes, I know how to do that geometric progression and I can do it with 2^(-k). But I have also tried to differentiate without success. Could you please guide me? Thank you!

5. ## Re: convergence problem

Originally Posted by mathfn
Yes, I know how to do that geometric progression and I can do it with 2^(-k). But I have also tried to differentiate without success. Could you please guide me? Thank you!
As I recall the series is $\displaystyle \sum\limits_{k = 1}^\infty {\frac{k}{{{2^k}}}}$.
Lets consider $\displaystyle y = \sum\limits_{k = 1}^\infty {x^k}=\frac{x}{1-x}$ where $|x|<1$
Now
$\displaystyle \frac{dy}{dx} = \sum\limits_{k = 1}^\infty {kx^{k-1}}=\frac{1}{(1-x)^2}$, thus $\displaystyle x\cdot \frac{dy}{dx} = \sum\limits_{k = 1}^\infty {kx^{k}}=\frac{x}{(1-x)^2}$
So use $x=\dfrac{1}{2}$ to get $\displaystyle \sum\limits_{k = 1}^\infty {\frac{k}{{{2^k}}}}=~?$

P.S. SEE HERE