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Let $a_n $ be a sequence of real numbers. Consider the infinite product

$$prodlimits_n=1^infty frac 1 1+a_n.$$

When do we say that the above infinite product is absolutely convergent? In the notion of an infinite series I know that a series $sumlimits_n=1^infty a_n$ is absolutely convergent if $sumlimits_n=1^infty |a_n|$ is convergent. For the case of infinite series I also know that absolute convergence implies convergence. Do all these results hold for infinite product too? Please help me in this regard.

Thank you so much for your valuable time.

For convergence of the product, we need $limlimits_ntoinftya_n=0$. That means that there is an $N$ so that for $nge N$, $|a_n|lt1$. Then,
$$
logleft(prod_n=N^inftyfrac11+a_nright)=-sum_n=N^inftylog(1+a_n)
$$
Thus, the convergence of the product is the same as the convergence of the sum of the logs. Absolute convergence of the product is the same as absolute convergence of the sum of the logarithms.

Therefore, absolute convergence of the product implies convergence of the product.

Definition

The infinite product $prod (1+b_j)$ is said to converge absolutely iff the infinite series $sum b_j$ converges absolutely.

(Any text that discusses infinite products should include this definition. And it should include the theorem that if an infinite product converges absolutely, then it converges.)

Your case:
$$
prodfrac11+a_j = prod (1+b_j),
$$
where

$$
b_j = 1 - frac11+a_j = fraca_j1+a_j
$$

So the criterion becomes:

$$
sumleft|fraca_j1+a_jright| < +infty
$$
As noted, this is equivalent to $sum |a_j| < +infty$.

If we define $$S=prod_n=1^inftyleft|1over 1+a_nright|$$then we have $$ln S=sum_n=1^inftyln left|1over 1+a_nright|$$Since boundedness of $S>0$ is equivalent to the boundedness of $ln S$, then any criteria on the absolute convergence of product of $mathbfleft1over 1+a_nright$ is equivalent to that for the convergence of the sum of $mathbflnleft$.