Convergence of Squared Series Proof for Positive Sequences

[dancergirlie]

Homework Statement

Assume that the series(an) is convergent and that an >= 0 for all n in N. Prove that the series((a^2)n) converges.

Homework Equations

The Attempt at a Solution

Alright, this is what I've got so far:
Assume that the series of an is convergent and that an>=0 for all n in N. In order for the series to be convergent, that would mean that the sequence (an) converges to 0. By definition of convergence, that would mean for epsilon greater than 0, there exists an N in N so that for n>=N:
|an|<epsilon and furthermore:
-epsilon<an<epsilon

This is where I get stuck... am I allowed to just multiply through by an to show that
-e(an)<((a^2)n)<e(an)?

And since series of (an) converges, and e is a constant, that would mean that series(epsilon*an) also converges, and by the comparison test that would mean that:
series((a^2)n) converges as well.

I don't know if what I'm doing is right, if it isn't then any tips would be great!

 

[latentcorpse]

why must a_n converge to 0?

consider the sequence a_n=\{1,2,3,4,5,5,5,5,5,5,5,5,5,5,5,...\}. here you have a_n \geq 0 and it converges to 5 \neq 0

i think your sort of on the right track though.
im not very good at sequences so i would wait for someone else's opinion but surely you could say that since a_n converges, \exists N_0 \in \mathbb{N} such that \forall n \geq N_0 we have -\epsilon &lt; a_n &lt; \epsilon
now multiply through by a_n
getting -\epsilon a_n &lt; (a_n)^2 &lt; \epsilon a_n
now the sequence -\epsilon a_n will be bounded as follows from the convergence of a_n:
-\epsilon ( -\epsilon) &lt; -\epsilon a_n &lt; -\epsilon (\epsilon)
and similarly for the sequence \epsilon a_n:
\epsilon (-\epsilon) &lt; \epsilon a_n &lt; \epsilon(\epsilon)
so (a_n)^2 is bounded between
-\epsilon(-\epsilon) &lt; (a_n)^2 &lt; \epsilon( \epsilon) \Rightarrow \epsilon^2 &lt; (a_n)^2 &lt; \epsilon^2
then by the sandwich theorem a_n converges.

however, as i said, I am not very good at analysis so this could be completely wrong but its my shot at the answer. wait for someone better to give their 2 cents...

 

[Dick]

Since you know a_n->0 since the series a_n converges, then there is an N such that |a_n|<1 for all n>N. That means |a_n|^2<|a_n|. Think comparison test.

 

[dancergirlie]

so basically what you did was the same as me but you assumed that epsilon was less than or equal to 1, but you still multiplied through by a_n? If that's the case, then I understand, if not, please correct me where I'm wrong...

 

[dancergirlie]

but i do understand for all values between 0 and 1, the square of that value is less than the original value, would it be better to just explain that rather than multiplying through by a_n?

 

[Dick]

but i do understand for all values between 0 and 1, the square of that value is less than the original value, would it be better to just explain that rather than multiplying through by a_n?

You can do it either way, sure. Showing |a_n|^2<e|a_n| also shows |a_n|^2 converges. Just seemed nicer to pick e=1.

 

[dancergirlie]

alright, thanks so much for your help!

 

[latentcorpse]

why does this ahve to converge to 0?

 

[VeeEight]

http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx
See theorem halfway down the page

 

[dancergirlie]

Looking at your example, saying that an converges to 5 would mean that the series of an, would be increasing by 5 each time. The only way for the series to approach a number is if the values of an you keep adding on get smaller and smaller, and eventually approach zero, that way the sum can approach a specific number. Meaning if an doesn't approach zero that means that the sum of an diverges, because it would be unbounded.