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

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    Kinda looks like [this](https://en.wikipedia.org/wiki/Euler_product) – 2017-01-24
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    For the second question, it is a special value of the [*Euler function*](https://en.wikipedia.org/wiki/Euler_function), and I suspect that your product has no known closed. – 2017-01-24
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    The first one is rather obvious: $(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})\cdots = \phi(\frac{1}{2})$. The second one is less obvious, but the following trick does the job: $$ \prod_{k=1}^{\infty} (1 + 2^{-k}) = \prod_{k=1}^{\infty} \frac{1-2^{-2k}}{1-2^{-k}} = \frac{\phi(\frac{1}{4})}{\phi(\frac{1}{2})}. $$ Of course, without knowledge on the Euler function, this is nothing but a reformulation of the problem. – 2017-01-24
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    Oops. Thanks. Like it... – 2017-01-24
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    For the record, $\phi(\frac{1}{2}) \approx 0.2887880951...$, and $\phi(\frac{1}{4})/\phi(\frac{1}{2}) \approx 2.3842310290...$. – 2017-01-24

2 Answers 2

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HINT:

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

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

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    The second one doesn't look pretty much like an improvement. +1 for the first one, though. – 2017-01-24
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    Oops! Pretty easy... What about the second one? How to simplify it? – 2017-01-24
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$$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})(1-\frac{1}{6})(1+\frac{1}{7})... $$ 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

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

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    Note that the term $(1 + \frac{1}{1})$ is "missing" from the front of the infinite product; if this term was present, it would cancel out with $(1 - \frac{1}{2})$ in the same way. – 2017-01-24
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    @ Michael Seifert : Well observed ! Then, the result would be $=1$. – 2017-01-24