Prove that $\left(1-\frac{1}{2}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\dots$ is convergent 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?
 A: From $$\sum_{n=1}^{\infty}\ln \left(1+\frac{(-1)^n}{2^n}\right)$$
you can use the alternating series test. Specifically, the sum is equal to $$\sum_{n=1}^{\infty}\left(-1\right)^{n}\ln\left(\left(1+\frac{\left(-1\right)^{n}}{2^{n}}\right)^{\left(-1\right)^{n}}\right)$$
Then it is enough to show that $$\ln\left(\left(1+\frac{\left(-1\right)^{n}}{2^{n}}\right)^{\left(-1\right)^{n}}\right) > \ln\left(\left(1+\frac{\left(-1\right)^{n+1}}{2^{n+1}}\right)^{\left(-1\right)^{n+1}}\right)$$
Exponentiating both sides yields $$\left(1+\frac{\left(-1\right)^{n}}{2^{n}}\right)^{\left(-1\right)^{n}} > \left(1+\frac{\left(-1\right)^{n+1}}{2^{n+1}}\right)^{\left(-1\right)^{n+1}}$$
Split this into two cases: $n$ odd and $n$ even. For $n$ odd, the inequality would simplify to $$\left(1-\frac{1}{2^{n}}\right)^{-1} > \left(1+\frac{1}{2^{n+1}}\right)^{1}$$
This is easy to show since $\left(1-\frac{1}{2^{n}}\right)\left(1+\frac{1}{2^{n+1}}\right) < 1-\frac{1}{2^{n+1}} < 1$. Similarly for $n$ even, the inequality would be $$\left(1+\frac{1}{2^{n}}\right)^{1} > \left(1-\frac{1}{2^{n+1}}\right)^{-1}$$
This is also easy to show since $\left(1+\frac{1}{2^{n}}\right)\left(1-\frac{1}{2^{n+1}}\right) = 1+\frac{1}{2^{x}}\left(\frac{1}{2}-\frac{1}{2^{\left(n+1\right)}}\right) > 1$. Therefore by the alternating series test, the sum would converge.
A: Hint: it suffices to show the product is absolutely convergent. In this case, we can use the fact that if $a_k>0$, $\prod_{k\ge 1}1+a_k$ converges iff $\sum_{k\ge 1}a_k$ converges.
A: 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 $2\ln(1/2)-1$.
