Convergence of Infinite Product

[holomorphic]

Homework Statement

If \sum a_{j} converges absolutely, and a_{j}\neq -1 for all j, then show \prod _{j=1} ^{\infty} (1+a_{j})\neq 0. Hint: Consider b_{j} such that (1+b_{j})(1+a_{j})=1. Show that \sum _{j=1} ^{\infty} b_{j} converges absolutely, and consider \prod _{j=1} ^{\infty} (1+a_{j}) \bullet \prod _{j=1} ^{\infty} (1+b_{j})

Homework Equations

The Attempt at a Solution

Taking the hint gives b_{j} = \frac{1}{1 + a_{j}} - 1, but I am not really sure how to show \sum b_{j} converges absolutely. I tried writing down inequalities I know, e.g. \left|a_{j} + 1 \right| \leq \left|a_{j}\right| + 1, and manipulating them to show that \sum \left|b_{j}\right| \leq \sum\left|a_{j}\right|... but it's not working. I also tried to write \sum \left|b_{j}\right| as a fraction with the product \prod (1+a_{j}) in the denominator, but the formula for the numerator turned out not to be so easy to write.

Any suggestions would be appreciated :)

 

[Dick]

bj=1/(1+aj)-1=(-aj)/(1+aj). lim aj->0. So for large enough j, |aj|<(1/2). Isn't that enough to show bj converges absolutely if aj does?

 

[holomorphic]

bj=1/(1+aj)-1=(-aj)/(1+aj). lim aj->0. So for large enough j, |aj|<(1/2). Isn't that enough to show bj converges absolutely if aj does?

I guess I don't understand why that's enough to show bj converges absolutely.

Ohh wait... so \left| b_{n} \right| \leq \left| a_{n} \right| when n>=N for some N and an converges absolutely, therefore bn converges absolutely. Right?

 

[Dick]

I guess I don't understand why that's enough to show bj converges absolutely.

Ohh wait... so \left| b_{n} \right| \leq \left| a_{n} \right| when n>=N for some N and an converges absolutely, therefore bn converges absolutely. Right?

I wouldn't say |b_n|<=|a_n|. a_n isn't necessarily positive. So |1+a_n| isn't greater than one. But you can show |b_n|<=C*|a_n| for some constant C.

 

[holomorphic]

I wouldn't say |b_n|<=|a_n|. a_n isn't necessarily positive. So |1+a_n| isn't greater than one. But you can show |b_n|<=C*|a_n| for some constant C.

So, supposing |a_n| < 1/2, then \left| b_{n} \right| = \frac{\left| a_{n} \right|}{\left| 1 + a_{n} \right|} &lt; 1, so that |b_n| < 2*|a_n| ?

If this is right, then thanks very much for your help.

 

[Dick]

So, supposing |a_n| < 1/2, then \left| b_{n} \right| = \frac{\left| a_{n} \right|}{\left| 1 + a_{n} \right|} &lt; 1, so that |b_n| < 2*|a_n| ?

If this is right, then thanks very much for your help.

Right.