Sum: $\sum_{n=1}^{\infty}\ln\left(1+\frac{8}{n^2+9n}\right)$ I have a problem with this sum, I've been trying to solve it for a while now without success.
$$S=\sum_{n=1}^{\infty}\ln\left(1+\frac{8}{n^2+9n}\right)$$
I tried this which I found in my texbook which seems relative, but Im not sure how to apply it to the problem:
$$a_n=f(n)-f(n+k) \  ,\forall n\geq1$$
 A: Hint: 
$$\ln\left(1+\frac{8}{n^2+9n}\right)=\ln\left(\frac{n^2+9n+8}{n^2+9n}\right)=\ln\left({n^2+9n+8}\right)-\ln\left({n^2+9n}\right)$$
A: $$a_n =\ln \left(1+\dfrac8{n^2+9n}\right) = \ln \left(\dfrac{n^2+9n+8}{n^2+9n}\right) = \ln (n+1) - \ln(n) + \ln(n+8) - \ln(n+9)$$
Hence,
$$a_n = b_{n+1} - b_n$$
where
$$b_n = \ln(n) - \ln(n+8)$$
A: Finally, 
\begin{eqnarray}
&&\sum\limits_{n = 1}^\infty  {\ln \left( {1 + \frac{8}{{{n^2} + 9n}}} \right)}   \\
&=& \sum\limits_{n = 1}^\infty  {\ln \frac{{n + 1}}{n}}  - \sum\limits_{n = 1}^\infty  {\ln \frac{{n + 8}}{{n + 9}}}  \\
&=& \ln \prod\limits_{n = 1}^\infty  {\frac{{n + 1}}{n}}  - \ln \prod\limits_{n = 1}^\infty  {\frac{{n + 8}}{{n + 9}}}  \\
&=& \ln \prod\limits_{n = 1}^8 {\frac{{n + 1}}{n}}  + \ln \prod\limits_{n = 9}^\infty  {\frac{{n + 1}}{n}}  - \ln \prod\limits_{n = 1}^\infty  {\frac{{n + 8}}{{n + 9}}}   \\
&=& \ln \prod\limits_{n = 1}^8 {\frac{{n + 1}}{n}}  + \ln \prod\limits_{n = 1}^\infty  {\frac{{n + 9}}{{n + 8}}}  - \ln \prod\limits_{n = 1}^\infty  {\frac{{n + 8}}{{n + 9}}}   \\
&=& \ln \prod\limits_{n = 1}^8 {\frac{{n + 1}}{n}}  \\
&=& 2\ln 3 
\end{eqnarray}
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\color{#66f}{\large S}&=\sum_{n = 1}^{\infty}\ln\pars{1 + {8 \over n^2+9n}}
=8\sum_{n = 1}^{\infty}\int_{0}^{1}{\dd x \over n^{2} + 9n + 8x}
=8\int_{0}^{1}\sum_{n = 1}^{\infty}{1 \over n^{2} + 9n + 8x}\,\dd x
\end{align}

The sum over $\ds{n}$ can be performed by " standard means ". Upon integration the result is
  $\begin{array}{|c|}\hline\color{#66f}{\large 2\ln\pars{3} \approx 2.1972}\\ \hline\end{array}$.

