For an infinite product $\prod a_k$ to converge we need
- at most finitely many zero factor, let be $m$ the maximum index of them
- $c=\lim_{n\to \infty}\prod_{k=m+1}^n a_k$ must exists, and
- $c\ne 0$.
My question is "Why the additional condition 3?"
Consider
$$\tag{1}
\prod_{k=1}^\infty \frac{n}{n+1}=\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots
$$
The $n$th partial product would be $1/n$, thus the limit is zero.
The definition above excludes (1) from the converging infinite products, but I do not understand what is bad about (1) converging to zero. There must be some consideration behind it.
EDIT
My question could be read as follows: What are the advantages of this definition? Is there a better (easier to develop by excluding the zero) definition ? Why is the zero limit excluded (even in case there are no zero factors)?
Best Answer
Yes, this is the definition in all books covering infinite products. We say $\prod \frac{n}{n+1}$ "diverges to $0$", and is not included when we say an infinite product "converges". The reason for this definition is that it is useful, for example in complex analysis.
One example: (there are many others) $$ \sin z= z \prod_{n=1}^\infty \left(1-{\frac {z^{2} } {\pi^2n^{2} } } \right) $$ where (for all complex $z$) it is a convergent infinite product. Therefore, we may read the zeros of $\sin$ from it directly. With infinite products that possibly diverge to $0$, we cannot do that.
As all mathematics students know, the principle $ab = 0 \Longrightarrow (a=0\text{ or }b=0)$ is very useful. We want to keep this useful fact for infinite products also! The is the first (and simplest, of many) reasons that convergence of infinite products is defined this way.