How to sum up the following sequence The following summation arises is a Bayesian model where Beta distribution is used as the prior for Negative Binomial distribution. The summation is used to assess the risk of the resulting estimator. However, to this stage it is only a pure mathematical problem. My question is how to find the following sum, please? Thank you!
\begin{align*}
\sum_{n=2}^\infty \frac{(n+1)(n-1)}{(n+7)^2(n+8)} (1-\theta)^{n-2}.
\end{align*}

First,
$${{\left(n-1\right)\,\left(n+1\right)}\over{\left(n+7\right)^2\,
 \left(n+8\right)}}={{63}\over{n+8}}-{{62}\over{n+7}}+{{48}\over{\left(n+7\right)^2}}$$
Thus, with $\xi=1-\theta$
$$f(\theta)=\sum_{n=2}^\infty \frac{(n+1)(n-1)}{(n+7)^2(n+8)} (1-\theta)^{n-2}\\=
63 \sum_{n=2}^\infty \frac{(1-\theta)^{n-2}}{n+8}
-62\sum_{n=2}^\infty \frac{(1-\theta)^{n-2}}{n+7}
+48 \sum_{n=2}^\infty \frac{(1-\theta)^{n-2}}{(n+7)^2}
\\=
63 \frac{1}{\xi^{10}} \sum_{n=10}^\infty \frac{\xi^{n}}{n}
-62 \frac{1}{\xi^{9}} \sum_{n=9}^\infty \frac{\xi^{n}}{n}
+48 \frac{1}{\xi^{9}} \sum_{n=9}^\infty \frac{\xi^{n}}{n^2}
$$
And
$$\sum_{n=1}^\infty \frac{\xi^n}{n}=-\log (1-\xi)=-\log \theta$$
$$\sum_{n=1}^\infty \frac{\xi^n}{n^2}=\mathrm{Li_2}(\xi)$$
Where $\mathrm{Li_2}$ is the dilogarithm.
Therefore,
$$f(\theta)=-63\frac{\log \theta}{(1-\theta)^{10}}+62\frac{\log \theta}{(1-\theta)^{9}}+48\frac{\mathrm{Li_2}(\theta)}{(1-\theta)^{9}}-g(\theta)$$
With
$$g(\theta)=63 \sum_{n=1}^{9} \frac{(1-\theta)^{n-10}}{n}
-62  \sum_{n=1}^{8} \frac{(1-\theta)^{n-9}}{n}
+48  \sum_{n=1}^{8} \frac{(1-\theta)^{n-9}}{n^2}$$
