if An>=0 and summation(An) converge, then summation ((An)^2) also converge.please...
This is a corollary of a more comprehensive theorem. If $\displaystyle \sum_{n\in\mathbb{N}}a_n$ converges absolutely then $\displaystyle \sum_{n\in\mathbb{N}}a_n^2$ converges as well.
To see this merely note that since $\displaystyle a_n\to0$ there exists some $\displaystyle N\in\mathbb{N}$ such that $\displaystyle N\leqslant n\implies |a_n|<1\implies a_n^2<|a_n|$ and so $\displaystyle \sum_{j=N}^{\infty}a_n^2<\sum_{j=N}^{\infty}|a_n|$ and the conclusion follows.