Conv. Conditionally Proff

[DEMJ]

Homework Statement

If $$a_k$$ is decreasing and it's limit is 0 as $$k \to \infty$$ and $$\sum_{k+1}^{\infty} b_k$$ converges conditionally, then $$\sum_{k=1}^{\infty} a_k b_k$$ converges

Homework Equations

This is true or false.

The Attempt at a Solution

I think it is false because if we let $$a_k = \frac{1}{\sqrt{k}}, b_k= \frac{(-1)^k}{\sqrt{k}}$$ we satisfy our initial conditions but $$a_k \cdot b_k = \frac{1}{k}$$ so $$\sum_{k=1}^{\infty} a_k b_k$$ diverges.
Is this correct?

 

[lanedance]

i think you missed a (-1)^k in the a_k term?

 

[DEMJ]

that would make $$a_k b_k = \frac{(-1)^{2k}}{k}$$ which is convergent, I think. I meant what I put but apparently it does not work?

 

[lanedance]

isn't it
$$a_k b_k = \frac{(-1)^{2k}}{k} = \frac{((-1)^2)^k}{k} = \frac{1}{k}$$

i wasn't sure where the alternating negative went in your 1st post...

 

[DEMJ]

isn't it
$$a_k b_k = \frac{(-1)^{2k}}{k} = \frac{((-1)^2)^k}{k} = \frac{1}{k}$$

i wasn't sure where the alternating negative went in your 1st post...

You are right, I owned my self by basic algebra :tongue2: