Multiplication is hard to work with, we can transform it into summation
$$ \begin{align*} \left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\dots\left(1+\dfrac{1}{a}\right)&=e^{\ln(1+1)+\ln(1+1/2)+\ln(1+1/3)+\dots+\ln(1+1/a)}\\ &=e^{\sum^a_{n=1}\ln\left(1+\frac{1}{n}\right)} \end{align*} $$
Which means if \(\displaystyle\sum^a_{n=1}\ln\left(1+\frac{1}{n}\right)\) converges as \(a\to\infty\), then \(\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\dots\) also converges.
We are considering it as a finite series here, because it yields interesting results.
$$ \begin{align*} \sum^a_{n=1}\ln\left(1+\frac{1}{n}\right)&=\sum^a_{n=1}\ln\left(\frac{n+1}{n}\right)\\ &=\sum^a_{n=1}(-\ln n+\ln(n+1))\\ &=-\ln 1+\ln 2-\ln 2+\ln 3-\dots+\ln a\\ &=\ln a \end{align*} $$
In other words, \(\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\dots\left(1+\dfrac{1}{a}\right)=e^{\ln a}=a\). Which is really cool, as the multiplication series(?) evaluates to its length.
Nonetheless, for such series with inifnite length, its value will also evaluate to infinite, therefore it does not converge.