prove this

**nzm88** · February 8th 2010, 07:59 PM

if An>=0 and summation(An) converge, then summation ((An)^2) also converge.please...

**Drexel28** · February 8th 2010, 08:12 PM

Originally Posted by nzm88

if An>=0 and summation(An) converge, then summation ((An)^2) also converge.please...

This is a corollary of a more comprehensive theorem. If $\sum_{n\in\mathbb{N}}a_n$ converges absolutely then $\sum_{n\in\mathbb{N}}a_n^2$ converges as well.

To see this merely note that since $a_n\to0$ there exists some $N\in\mathbb{N}$ such that $N\leqslant n\implies |a_n|<1\implies a_n^2<|a_n|$ and so $\sum_{j=N}^{\infty}a_n^2<\sum_{j=N}^{\infty}|a_n|$ and the conclusion follows.

**Bruno J.** · February 8th 2010, 08:12 PM

Well from some point on, say for $n>n_0$ , we will have $|A_n|<1$ . But then $A_n^2<|A_n|$ ; so just compare the tails of the two series.

**Bruno J.** · February 8th 2010, 08:13 PM

Great minds think alike

**Drexel28** · February 8th 2010, 08:14 PM

Originally Posted by Bruno J.

Great minds think alike

Agreed. We'll probably be in a think-tank someday together. We can then go to bars and try to impress women with our bridling intellect and manish good looks.

**Bruno J.** · February 8th 2010, 08:20 PM

It'd be an honour to wing-man you!

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