$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})...$

Question

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 $$(1+\frac{1}{2})(1+\frac{1}{4})(1+\frac{1}{8})(1+\frac{1}{16})...$$

Answer 1

HINT:

For the first question, $$\left(1+\dfrac1{2n+1}\right)\left(1-\dfrac1{2n+2}\right)=1$$

$$\prod_{r=1}^n\left(1-\dfrac1{2^r}\right)=\dfrac{\prod_{r=1}^n(2^r-1)}{2^{1+2+\cdots+n}}$$

Answer 2

$$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})(1-\frac{1}{6})(1+\frac{1}{7})... $$ 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

$$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})(1-\frac{1}{6})(1+\frac{1}{7})... = \frac{1}{2}$$