1. ## complex analysis series

Let $a_n, b_n$ be nonzero complex numbers. Suppose $lim_{n\to\infty}|\frac{a_n}{b_n}|=l$ exists, and $l \not= 0, \infty$.
1. Show that if one of the series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$ converges absolutely, then so does the other.
2. What if $l=0$ or $l=\infty$?

For #1, I assume $\sum_{n=0}^\infty a_n$ converges absolutely and trying to prove $\sum_{n=0}^\infty b_n$ converges absolutely. That is, $\forall \epsilon>0$ there exists $N \in \mathbb{N}$ such that $|b_{n+1}|+|b_{n+2}|+...+|b_{n+p}|<\epsilon$ for $p>0$.
I tried to use what I have to prove this, but not getting what I want. Can I get some help?

I don't know #2...

2. Originally Posted by dori1123
Let $a_n, b_n$ be nonzero complex numbers. Suppose $lim_{n\to\infty}|\frac{a_n}{b_n}|=l$ exists, and $l \not= 0, \infty$.
1. Show that if one of the series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$ converges absolutely, then so does the other.
2. What if $l=0$ or $l=\infty$?

For #1, I assume $\sum_{n=0}^\infty a_n$ converges absolutely and trying to prove $\sum_{n=0}^\infty b_n$ converges absolutely. That is, $\forall \epsilon>0$ there exists $N \in \mathbb{N}$ such that $|b_{n+1}|+|b_{n+2}|+...+|b_{n+p}|<\epsilon$ for $p>0$.
I tried to use what I have to prove this, but not getting what I want. Can I get some help?

I don't know #2...

As $\lim_{n\to\infty}\frac{a_n}{b_n}=l$ , we have that $|l|-\epsilon<\left|\frac{a_n}{b_n}\right|<|l|+\epsilon$ $\Longleftrightarrow (|l|-\epsilon)|b_n|<|a_n|<(|l|+\epsilon)|b_n|$ for all but a finite number of indexes $n\in\mathbb{N}$ , so:

1) if $\sum\limits_{n=0}^\infty |a_n|$ converges then so converges $\sum\limits_{n=0}^\infty (|l|-\epsilon)|b_n|$ by the comparison test and thus...

2) if $\sum\limits_{n=0}^\infty |b_n|$ converges then again by the comparison test converges the series ....

End both arguments above, and note that we can choose $\epsilon > 0$ so as to get $|l|-\epsilon > 0$ (why is this important?)

Tonio