Let and be an absolutely convergent series of real numbers. Prove that the series converges.

Now by the definition of absolute convergence we know that:

for some real number and for some real number .

So we have .

How do I tie in square roots?

Printable View

- Oct 1st 2013, 11:26 AMhughmistlerAbsolute Convergence
Let and be an absolutely convergent series of real numbers. Prove that the series converges.

Now by the definition of absolute convergence we know that:

for some real number and for some real number .

So we have .

How do I tie in square roots? - Oct 1st 2013, 01:11 PMShakarriRe: Absolute Convergence
- Oct 1st 2013, 01:44 PMPlatoRe: Absolute Convergence
- Oct 1st 2013, 07:58 PMbobcantor1983Re: Absolute Convergence
- Oct 1st 2013, 08:10 PMArythRe: Absolute Convergence
- Oct 1st 2013, 08:46 PMbobcantor1983Re: Absolute Convergence
I understand how you can do that with addition and that the sum of two convergent series converges, but how can that apply to multiplying two convergent series?

- Oct 1st 2013, 09:00 PMArythRe: Absolute Convergence
- Oct 2nd 2013, 01:00 PMbobcantor1983Re: Absolute Convergence
__So let me put this together, does this all make sense, am I missing anything?__

Suppose and are absolutely convergent series of real numbers. We need to show that converges. We will show this using the comparison test.

The comparison test states: Suppose that we have two series, and , with for all and for all ; then if is convergent so is .

Now by the definition of absolute convergence we know that:

for some real number and for some real number .

Let us consider that

We show this is true:

Now consider that,

**Since we have shown that every element is less than or equal to this new series, and it is clear that this new series converges since it is the sum of two convergent series multiplied by a constant. Hence by the comparison test, is convergent.** - Oct 2nd 2013, 03:24 PMPlatoRe: Absolute Convergence