Probability of ultimate extinction? Need to show that an infinite series is less than $1$ I have the following probability generating function for a branching process -
$$G_n(s) = \frac{n}{n+1} + \sum_{r=1}^{\infty}\frac{n^{r-1}}{(n+1)^{r+1}}s^r$$
It says in a book that extinction is certain for this process. Which implies that the mean of the probability generating function must be $\le1$.
I.e. $G_n'(1) \le 1$
But
$$G_n'(1) = \sum_{r=1}^{\infty}\frac{n^{r-1}}{(n+1)^{r+1}}$$
and I don't see how to show this is less than or equal to $1$?
 A: If
$G_n'(s)
=\sum_{r=1}^\infty \frac{n^{r-1}}{(n+1)^{r+1}}\frac{d}{dx}\left[s^r\right]
= \sum_{r=1}^\infty\frac{rn^{r-1}}{(n+1)^{r+1}}s^{r-1}
$,
then
$G_n'(1)
= \sum_{r=1}^\infty\frac{rn^{r-1}}{(n+1)^{r+1}}
= \frac1{n(n+1)}\sum_{r=1}^\infty\frac{rn^{r}}{(n+1)^{r}}
= \frac1{n(n+1)}\sum_{r=1}^\infty rz^r
$,
where
$z = \frac{n}{n+1}
$.
Since
$\sum_{r=1}^\infty rz^r
= \frac{z}{(1 - z)^2}
$,
$\begin{array}\\
G_n'(1)
&=\frac1{n(n+1)}\frac{z}{(1 - z)^2}\\
&=\frac1{n(n+1)}\frac{\frac{n}{n+1}}{(1 - \frac{n}{n+1})^2}\\
&=\frac{\frac{1}{(n+1)^2}}{\frac{1}{(n+1)^2}}\\
&=1\\
\end{array}
$
More generally,
$\begin{array}\\
G_n'(s)
&= \sum_{r=1}^\infty\frac{rn^{r-1}}{(n+1)^{r+1}}s^{r-1}\\
&= \frac1{sn(n+1)}\sum_{r=1}^\infty\frac{rn^{r}}{(n+1)^{r}}s^{r}\\
&= \frac1{sn(n+1)}\sum_{r=1}^\infty r\left(\frac{ns}{n+1}\right)^r\\
&= \frac1{sn(n+1)}\sum_{r=1}^\infty rz^r
\quad \text{ where } z=\frac{ns}{n+1}\\
&= \frac1{sn(n+1)}\frac{z}{(1-z)^2}\\
&= \frac1{sn(n+1)}\frac{\frac{ns}{n+1}}{(1-\frac{ns}{n+1})^2}\\
&= \frac1{(n+1)^2}\frac{1}{(1-\frac{ns}{n+1})^2}\\
&= \frac{1}{(n+1-ns)^2}\\
&= \frac{1}{(1+n(1-s))^2}\\
\end{array}
$
Note that if
$1+n(1-s)
= 0$,
or
$s
=1+\frac1{n}
$,
then
$G_n'(s)
= \infty
$.
A: Note that your series for $G_n$ is a geometric series.  You can find a simple closed form for the sum.
Then take the derivative.
