If Ʃa_n converges, with a_n > 0, then Ʃ(a_n)^2 always converges

[DotKite]

Homework Statement

if Ʃa_n converges, with a_n > 0, then Ʃ(a_n)^2 always converges

Homework Equations

n/a

The Attempt at a Solution

I am at a complete loss. I have tried using partial sums, cauchy criterion, and I tried using ratio test which seems to work but I am not sure.

Since Ʃa_n converges then by ratio test

lim n->∞ a_n+1 / a_n < 1

Now we apply ratio test to Ʃ(a_n)^2

lim n→∞ (a_n+1)^2 / (a_n)^2 = (lim n→∞ a_n+1 / a_n)^2 < 1^2 = 1

Thus by ratio test Ʃ(a_n)^2 converges.

This working did not utilize the condition a_n > 0, so it seems suspect to me.

 

[jbunniii]

Hint: if $0 < a_n < 1$, then $a_n^2 < a_n$.

 

[DotKite]

Hint: if $0 < a_n < 1$, then $a_n^2 < a_n$.

Ok so you use comparison test right? since (a_n)^2 < a_n and a_n converges then (a_n)^2 converges as well, but is it correct to assume 0<a_n<1? Can you assume that because since
Ʃa_n converges that implies a_n → 0 and for some n≥N a_n is in (0,1)?

 

[jbunniii]

Ok so you use comparison test right? since (a_n)^2 < a_n and a_n converges then (a_n)^2 converges as well, but is it correct to assume 0<a_n<1? Can you assume that because since
Ʃa_n converges that implies a_n → 0 and for some n≥N a_n is in (0,1)?

Yes, that's right, or putting it more precisely, there exists an $N$ such that $a_n \in (0,1)$ for all $n \geq N$. And then use the comparison test.

 

[DotKite]

thanks a bunch J