Does the infinite product $\frac{2}{1}\times\frac{4}{3}\times\frac{6}{5}\times\dots$ converge?

Question

I was wondering whether the infinite product

$$\prod_{i=0}^{\infty}\frac{2n + 2}{2n + 1} = \frac{2}{1}\times\frac{4}{3}\times\frac{6}{5}\times\dots$$

converges. For all I know it is certainly not absolutely convergent, but is it at least conditionally so? Telescoping is not possible in this case. I know that

$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \ln{2}$$

but if I take the logarithm off this infinite product, I'd get

$$ \ln{1} - \ln{\frac{1}{2}} + \ln{\frac{1}{3}} - \ln{\frac{1}{4}} + \dots$$

which is not conclusive.

Answer 1

Hint: If $a,b,c,\ldots$ are all positive $$(1+a)(1+b)(1+c)\cdots\ge 1+a+b+c+\cdots.$$