How to establish the convergence?

Question

How to show that $\prod\limits_{n=1}^\infty \frac{1+x^{\delta^n}}{2}$ is convergent if $x>0, 0<\delta<1$ ?

Thanks in advance

Answer 1

Does it work like this? \begin{align} \log\prod_{n=1}^\infty \frac{1+x^{\delta^n}}{2} &= \sum_{n=1}^\infty \log{\frac{1+x^{\delta^n}}{2}} \\ &= \sum_{n=1}^\infty\delta^n\log{\left(\frac{1+x^{\delta^n}}{2}\right)^{\delta^{-n}}} \end{align}

The expression $\log{\left(\frac{1+x^{\delta^n}}{2}\right)^{\delta^{-n}}}$ is majorised by a constant sequence if $x > 0$ since it converges monotonically so the whole summand is majorised by $\delta^n$ which is for the given $\delta\in(0,1)$ the geometric series. Convergence of the logarithm of the sequence is equivalent to the convergence of the sequence itself.