TITLE: Prove that $\left(1-\frac{1}{2}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\dots$ is convergent
QUESTION [5 upvotes]: Prove that $\left(1-\frac{1}{2}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\dots$ is convergent:
Let us rewrite this into
$$\prod_{n=1}^{\infty}\left(1+\frac{(-1)^n}{2^n}\right)$$
This product converges if
$$\sum_{n=1}^{\infty}\ln \left(1+\frac{(-1)^n}{2^n}\right)$$
does. I write now
$$\sum_{n=1}^{\infty}\ln \left(1+\frac{(-1)^n}{2^n}\right)<\sum_{n=1}^{\infty}\ln \frac{(-1)^n}{2^n}$$
I'm not sure about this step. Can someone help me with this please?

REPLY [2 votes]: By concavity of $\ln(x)$ we have $\ln(x)<x-1$ where $x-1$ is the tangent to our function at $x=1$. So firstly:
$$\sum_{n \text{ even}} \ln \left ( 1 + 2^{-n} \right ) < \sum_{n} \ln \left ( 1 + 2^{-n} \right ) < \sum 2^{-n}=1$$
This covers positive terms. For negative terms, we use Jensen's inequality, to show that between $x=1/2$ and $x=1$ our function lies above the line going through $(1/2, \ln(1/2))$ and $(1, 0)$. This gives $\ln(x) > 2\ln\left ( 1/2 \right ))\left ( 1-x \right )$ on this interval. As $\ln(x)$ is negative on $(1/2, 1)$, this implies:
$$\sum_{n\text{ odd}} \left |  \ln \left ( 1 - 2^{-n} \right ) \right |  < \sum_{n} \left |  \ln \left ( 1 - 2^{-n} \right ) \right | < 2\ln\left ( 1/2 \right )) \sum 2^{-n} = 2\ln(1/2)$$
Your sum is therefore absolutely convergent, specifically the original sum is bounded by $1-2\ln(1/2)$ and $2\ln(1/2)-1$.