Prove existence of $n$ for an infinite product I've been studying a paper in which the author says:
Fix $n$ such that $m^n \prod_{j=1}^n \frac{j}{j+\delta} &gt; 1$, where $1&lt;m&lt;\infty$, and $\delta &gt;0$.
I seem not to be able to show why such $n$ must exist. I tried rewriting it this way:
$m^n \prod_{j=1}^n \frac{j}{j+\delta} =m^n \prod_{j=1}^n (1-\frac{\delta}{j+\delta})= m^n \exp\bigg[\sum_{j=1}^n \log(1-\frac{\delta}{j+\delta})\bigg] \geq m^n\bigg[1+\sum_{j=1}^n \log(1-\frac{\delta}{j+\delta})\bigg]$
but after that I'm stuck. In one part the author writes
$m^n \prod_{j=1}^n (1-\frac{\delta}{j+\delta}) \geq m^n e^{c_0} \exp\bigg[-\delta\sum_{j=1}^n \frac{1}{j+\delta}\bigg]$, where $c_0$ is some constant,
but then I still don't know how this guarantees that there exists $n$ such that $m^n \prod_{j=1}^n \frac{j}{j+\delta} &gt; 1$. Does anyone have any idea on how to prove it?
Find an $N$ such that $\frac{mN}{N+\delta}&gt;1+\eta$ for some $\eta&gt;0$. Then take $M$ such that $(1+\eta)^M&gt;\left(\prod_{j=1}^N\frac{j}{j+\delta}\right)^{-1}$. Then take $n=N+M$.
Find an $N$ such that $\frac{mN}{N+\delta}&gt;1+\eta$ for some $\eta&gt;0$. Then take $M$ such that $(1+\eta)^M&gt;\left(\prod_{j=1}^N\frac{j}{j+\delta}\right)^{-1}$. Then take $n=N+M$.