# Thread: Proof with absolute and conditional convergence

1. ## Proof with absolute and conditional convergence

hello, i've stumbled upon another proof. the problem i s:

Prove that if sigma ace of n converges absolutely, then sigma (ace of n)^2 also converges. Then show by giving a counterexample that sigma (ace of n)^2 need not converge if sigma ace of n is only conditionally convergent.

For the first part, can I simply state that if sigma ace of n converges absolutely, it is implied that it converges. If sigma ace of n converges, then the square also converges because p>2 (p-series)?

2. what is sigma ace?
sigma=summation I guess
and is ace $\displaystyle a_n$?

3. yes, sorry about that. i dont know how to code the math signs.

4. I never heard of ace b4. The counterexample is

$\displaystyle \sum_{n=1}^{\infty}{(-1)^n\over \sqrt n}$.

5. thank you! so does that mean that my proof works?

6. The proof of part 1 is...

Since $\displaystyle \sum_{n=1}^{\infty}|a_n|$ converges, we have $\displaystyle |a_n|\to 0$, so just find N where $\displaystyle |a_n|<1$, whenever $\displaystyle n \ge N$.

Then $\displaystyle \sum_{n=N}^{\infty}a_n^2=\sum_{n=N}^{\infty}|a_n|\ cdot |a_n|< \sum_{n=N}^{\infty}|a_n|<\infty$.

And of course $\displaystyle \sum_{n=1}^{\infty}a_n^2=\sum_{n=1}^{N-1}a_n^2+\sum_{n=N}^{\infty}a_n^2<\infty$.

7. Originally Posted by wyhwang7
thank you! so does that mean that my proof works?

nope