# Thread: sum of the product of convergent series

1. ## sum of the product of convergent series

I'm not really sure what further to do with arbitrary series.

Theorem one states that if {$\displaystyle a_n$} and {$\displaystyle b_n$} have limits $\displaystyle A$ and $\displaystyle B$ respectively, then {$\displaystyle a_nb_n$} has the limit $\displaystyle AB$.

2. Originally Posted by davismj

I'm not really sure what further to do with arbitrary series.

Theorem one states that if {$\displaystyle a_n$} and {$\displaystyle b_n$} have limits $\displaystyle A$ and $\displaystyle B$ respectively, then {$\displaystyle a_nb_n$} has the limit $\displaystyle AB$.
Consider that since $\displaystyle \sum_{n=1}^{\infty}a_n$ converges we must have that $\displaystyle \lim\text{ }\sqrt[n]{|a_n|}\leqslant 1$. And so $\displaystyle \lim\text{ }\sqrt[n]{|a_n\cdot z^n|}=|z|\lim\text{ }\sqrt[n]{a_n}\leqslant |z|<1$.

3. Originally Posted by Drexel28
Consider that since $\displaystyle \sum_{n=1}^{\infty}a_n$ converges we must have that $\displaystyle \lim\text{ }\sqrt[n]{|a_n|}\leqslant 1$. And so $\displaystyle \lim\text{ }\sqrt[n]{|a_n\cdot z^n|}=|z|\lim\text{ }\sqrt[n]{a_n}\leqslant |z|<1$.
Wouldn't it be easier to say that since |$\displaystyle a_n$| and |$\displaystyle z^n$| become arbitrarily close ($\displaystyle \varepsilon$) to 0, their product also becomes arbitrarily close to zero, and thus the series is Cauchy?

4. Originally Posted by davismj
Wouldn't it be easier to say that since |$\displaystyle a_n$| and |$\displaystyle z^n$| become arbitrarily close ($\displaystyle \varepsilon$) to 0, their product also becomes arbitrarily close to zero, and thus the series is Cauchy?
I am not sure what you're saying, but I feel as though you are not considering that just because the summand converges to zero does not mean the series converges.

5. Originally Posted by Drexel28
I am not sure what you're saying, but I feel as though you are not considering that just because the summand converges to zero does not mean the series converges.

Is this better. the basic assumption is that since $\displaystyle \sum a_n$ converges, for some N the sum from n to infinty is less than epsilon. Since $\displaystyle \sum|z^n|$ is a geometric sum, its sum also converges (although less than 1 is sufficient), and so the product of these two will be less than epsilon as well. Is that kinda sorta right for an undergraduate introduction to complex analysis course?

6. Originally Posted by davismj

Is this better. the basic assumption is that since $\displaystyle \sum a_n$ converges, for some N the sum from n to infinty is less than epsilon. Since $\displaystyle \sum|z^n|$ is a geometric sum, its sum also converges (although less than 1 is sufficient), and so the product of these two will be less than epsilon as well. Is that kinda sorta right for an undergraduate introduction to complex analysis course?
The product of two series isn't the series of the products . Look at polynomials for example

7. Originally Posted by Drexel28
The product of two series isn't the series of the products . Look at polynomials for example
Ugh. So in other words, it would get me dirty looks from a college algebra teacher, let alone complex analysis. I fail.

How about this: Since $\displaystyle |z| < 1, |z^n|a_n < a_n$ for all n.

8. Originally Posted by davismj
Ugh. So in other words, it would get me dirty looks from a college algebra teacher, let alone complex analysis. I fail.

How about this: Since $\displaystyle |z| < 1, |z^n|a_n < a_n$ for all n.

You're getting warmer, but you need absolute convergence.