Good lower bound on an infinite product

242 Views Asked by At

I am trying to find a good lower bound on

$\prod_{k = j}^\infty (1- C \cdot 2^{-k})$,

where $C$ is constant, in terms of $j$ that goes to $1$ as $j\rightarrow \infty$. Does anyone know of any formulas which might be useful?

Looking on Wolfram alpha I find a closed form involving the 1/2-Pochhammer function, but am not sure how to use this information.

1

There are 1 best solutions below

0
On BEST ANSWER

Let your product be $P(j,C)$. I'll assume $0 < C < 2^j$, so all 1- C 2^{-k} > 0$.

We have $\log P(j,C) = \sum_{k=j}^\infty \log (1 - C \cdot 2^{-k})$, so we want good lower bounds on $\log(1-t)$ for small $t > 0$. Since $\log(1-t)$ is concave, we could use $$ \log(1-t) > t \log(1-t_0)/t_0$$ Thus $$ \log P(j,C) > \sum_{k=j}^\infty 2^{j-k} \log(1 - C 2^{-j}) = 2 \log(1 - C 2^{-j})$$ or $$P(j,C) > (1 - C \cdot 2^{-j})^2$$