Does Convergence of \(\sum a_n\) Imply Convergence of \(\sum a_n^2\)?

[stukbv]

Homework Statement

My questions is that if \suman converges then does \suman² converge?

Homework Equations

The Attempt at a Solution

I have said that if an converges we can say that a_n+1/_an tends to a where a < 1.

Then we can say from this that (a_n+1)²/a_n² tends to a², since a < 1 , a²< 1 too, thus \suman² converges.

Is this right?!?
Thanks a lot!

 

[hunt_mat]

Looks right to me, at first glance.

 

[kru_]

Another way to tackle the problem is to observe that the sum of an is Cauchy, so the sum of an*an would also be Cauchy.

 

[micromass]

Also, you should use [ itex] instead of [ tex] now. Using [ tex] will start a new line...

 

[Dick]

You are trying to use the ratio test backwards. That doesn't work. Suppose an converges and the ratio converges to 1 or -1? Or doesn't even converge at all? What happens then? In fact, if an isn't absolutely convergent, then your statement isn't true. Suppose an=(-1)^n/sqrt(n)?

 

[Maxter]

If your terms are all positive, then you can use the comparison test since the square function is increasing for positive numbers.

As Dick said, it's not always true.