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 $$\left| b_{n} \right| = \frac{\left| a_{n} \right|}{\left| 1 + a_{n} \right|} < 1$$, 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 $$\left| b_{n} \right| = \frac{\left| a_{n} \right|}{\left| 1 + a_{n} \right|} < 1$$, so that |b_n| < 2*|a_n| ?

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

Right.