sum of the product of convergent series

**davismj** · January 26th 2010, 05:01 PM

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

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

**Drexel28** · January 26th 2010, 06:16 PM

Originally Posted by davismj

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

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

Consider that since $\sum_{n=1}^{\infty}a_n$ converges we must have that $\lim\text{ }\sqrt[n]{|a_n|}\leqslant 1$ . And so $\lim\text{ }\sqrt[n]{|a_n\cdot z^n|}=|z|\lim\text{ }\sqrt[n]{a_n}\leqslant |z|<1$ .

**davismj** · January 27th 2010, 09:41 AM

Originally Posted by Drexel28

Consider that since $\sum_{n=1}^{\infty}a_n$ converges we must have that $\lim\text{ }\sqrt[n]{|a_n|}\leqslant 1$ . And so $\lim\text{ }\sqrt[n]{|a_n\cdot z^n|}=|z|\lim\text{ }\sqrt[n]{a_n}\leqslant |z|<1$ .

Wouldn't it be easier to say that since | $a_n$ | and | $z^n$ | become arbitrarily close ( $\varepsilon$ ) to 0, their product also becomes arbitrarily close to zero, and thus the series is Cauchy?

**Drexel28** · January 27th 2010, 12:44 PM

Originally Posted by davismj

Wouldn't it be easier to say that since | $a_n$ | and | $z^n$ | become arbitrarily close ( $\varepsilon$ ) 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.

**davismj** · January 27th 2010, 06:20 PM

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 $\sum a_n$ converges, for some N the sum from n to infinty is less than epsilon. Since $\sum|z^n|$ 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?

**Drexel28** · January 27th 2010, 06:22 PM

Originally Posted by davismj

Is this better. the basic assumption is that since $\sum a_n$ converges, for some N the sum from n to infinty is less than epsilon. Since $\sum|z^n|$ 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

**davismj** · January 27th 2010, 06:25 PM

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 $|z| < 1, |z^n|a_n < a_n$ for all n.

**Drexel28** · January 27th 2010, 06:56 PM

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 $|z| < 1, |z^n|a_n < a_n$ for all n.

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

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