Question

Need help with series proof.
Statement below

Answer 1

Proof that if \[a _{n} \ge 0\] and \[\sum_{}^{} a _{n} \] converges then \[\sum_{}^{} a_{n}\] also converges.

Answer 2

I tried to mess around a bit with the idea that for \[\sum_{}^{} a_{n}\] to converge, an should be smaller then 1 absolute, and such its square should be even smaller. But I have no idea if this is correct.

Help would be much appreciated, before I assume wrong things

Answer 3

I don't understand the question as you typed it

Answer 4

prove that if\[\sum a_n\text{ convergest, then }\sum a_n\text{ also converges}\]

Answer 5

?

Answer 6

Was the second sum supposed to be \(\Large\rm \sum_{}^{} a_{n+1}\) maybe?

Answer 7

Oh damn. I see what happened. Messed up the equation. The second sum is supposed to be \[\sum_{}^{} a_{n}^{2}\]
So if \[\sum_{}^{} a_{n}\] converges with an bigger or equal to 0, show that \[\sum_{}^{} a_{n}^{2}\] converges

Answer 8

are these sums over infinity?

Answer 9

I think that is implied yeah, given the question is about convergence where a series is convergent if the series of its partial sums converges / aproaches a given number. Now the only way this happens is if \[\lim_{n \rightarrow \infty} a_{n} = 0\]

It makes perfect sense that \[a_{n}^2\] would be convergent aswell, but I'm having a tough time giving a good mathematical proof.

Answer 10

http://en.wikipedia.org/wiki/Convergent_series

Answer 11

yeah i feel the exact same way; it's intuitive, but not obvious how to prove

Answer 12

As n approaches infinity, \(\Large\rm a_n\) is approaching 0.
So as n approaches infinity, \(\Large\rm a_n^2\) is approaching 0 much faster.

But yes, how to be more rigorous about that D: lol
hmmm

Answer 13

How about using the comparison test?

Answer 14

that will certainly work for the case where a_n < 1

Answer 15

but for cases with a_n > 0, we don't have that \(0\le a_n^2\le a_n\), which is what we would need to use that

Answer 16

a_n > 1

Answer 17

maybe the limit comparison test

Answer 18

only problem is that the limit of the ratios seems to be zero, or whatever the limit of a_n n-> infty is

Answer 19

I know. So let's see what we have.
We got the theorem that says that if If Sum a_n converges, then lim a_n = 0 as n goes to infinity. The converse is however not true, and even if the limit of a_n goes to 0, the series may still diverge. See the Divergence test.

One hint may be that a_n >= 0. This means it is always positive, and alternating series will not matter. So we know that the series does not only converge, but that it must converge absolutely.

Answer 20

how about something like the following\[for~~a_n\ge0\\\left(\sum a_n\right)^2\le\sum a_n^2 \]

Answer 21

oops, that should be \(\ge\)

Answer 22

i guess that is only true for \(a_n\ge 1\), but between that and the comparison test for \(a_n<1\) we can prove it... i think

Answer 23

I think I got it, double check though?
Using the Ratio test:
\[\lim_{n \rightarrow \infty} a_{n+1} / a_n < 1\]
This must be since the series converges, and for a converging series n<1
\[\lim_{n \rightarrow \infty} a_{n+1}^2 / a_{n}^2 \]
\[= \lim_{n \rightarrow \infty} (a_{n+1} / a_{n})^2 \le \left| 1^2 \right| < 1 \]
And with the ratio test, <1 = convergent.

Answer 24

oh yeah, that looks much better than what i was going to do. I think it works :)