$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})...$ Is there any well-known value (Or Approximation) for this?
$$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})...$$
we know that it converges as $$\sum_{i=2}^{\infty}\frac{(-1)^{i+1}}{i}=ln2-1$$
So there is a trivial upper bound $\frac{2}{e}$ for it. Is there any better result? In addition is there any similar result for
$$(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})...$$
or
$$(1+\frac{1}{2})(1+\frac{1}{4})(1+\frac{1}{8})(1+\frac{1}{16})...$$

$$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})(1-\frac{1}{6})(1+\frac{1}{7})... $$
Except the first term, consider consecutive couples of terms :
$(1+\frac{1}{3})(1-\frac{1}{4})= (\frac{4}{3})(\frac{3}{4})= 1$
$(1+\frac{1}{5})(1-\frac{1}{6})= (\frac{6}{5})(\frac{5}{6})= 1$
$(1+\frac{1}{7})(1-\frac{1}{8})= (\frac{8}{7})(\frac{7}{8})= 1$
And so on
All couples $=1$ . So, only the first term remains : $(1-\frac{1}{2})=\frac{1}{2}$
Finally
$$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})(1-\frac{1}{6})(1+\frac{1}{7})... = \frac{1}{2}$$
