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 $1$?
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 $.