How Can We Modify Convergent Series to Make Their Product Diverge?

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Homework Statement


Find a_n & b_n such that \sum_{n=0}^{\infty}a_n & \sum_{n=0}^{\infty}b_n are convergent series, but \displaystyle \sum_{n=0}^{\infty} \left( \sqrt{a_n} \cdot b_n \right) diverges.

Homework Equations


None.

The Attempt at a Solution


Try too hard for this but still cannot find such a_n & b_n.
 
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AfterSunShine said:

Homework Statement


Find a_n & b_n such that \sum_{n=0}^{\infty}a_n & \sum_{n=0}^{\infty}b_n are convergent series, but \displaystyle \sum_{n=0}^{\infty} \left( \sqrt{a_n} \cdot b_n \right) diverges.

Homework Equations


None.

The Attempt at a Solution


Try too hard for this but still cannot find such a_n & b_n.

That is not an acceptable post. You must show your efforts before we can offer tutorial help. Show us what you have tried so far please...
 
There is nothing to show here
basically I tried a_n = 1/n^2 & b_n = (-1)^n / n but failed
tried a_n = 1/n^2 & b_n = arctan (1/n) but failed
and so on...
 
AfterSunShine said:
basically I tried a_n = 1/n^2 & b_n = (-1)^n / n but failed
That is not so bad as a start. Can you modify the series in order to keep them converging, but doing so significantly slower?
And then you'll need some trick to get the product diverging - something that changes the sign flip thing...
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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