Does the Series Converge with Conditionally Convergent Multipliers?

[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