37. Compute
$$
\sum_{n=1}^{\infty} \frac{2 n+5}{2^{n} \cdot\left(n^{3}+7 n^{2}+14 n+8\right)}
$$
Answer: $\frac{137}{24}-8 \ln 2$
Solution: First, we manipulate using partial fractions and telescoping:
$$
\begin{aligned}
\sum_{n=1}^{\infty} \frac{2 n+5}{2^{n} \cdot\left(n^{3}+7 n^{2}+14 n+8\right)} & =\frac{1}{2} \cdot \sum_{n=1}^{\infty} \frac{1}{2^{n}}\left(\frac{2}{n+1}-\frac{1}{n+2}-\frac{1}{n+4}\right) \\
& =\frac{1}{4}-\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{2^{n} \cdot(n+4)}
\end{aligned}
$$

Now, consider the function $f(r, k):=\sum_{n=1}^{\infty} \frac{r^{n}}{n^{k}}$. We have
$$
\begin{aligned}
\frac{\partial f(r, k)}{\partial r} & =\frac{\partial}{\partial r} \sum_{n=1}^{\infty} \frac{r^{n}}{n^{k}}=\sum_{n=1}^{\infty} \frac{\partial}{\partial r}\left[\frac{r^{n}}{n^{k}}\right]=\sum_{n=1}^{\infty} \frac{r^{n-1}}{n^{k-1}}=\frac{1}{r} f(r, k-1) \\
\frac{d f(r, 1)}{d r} & =\frac{1}{r} \sum_{n=1}^{\infty} \frac{r^{n}}{n^{0}}=\frac{1}{r} \cdot \frac{r}{1-r}=\frac{1}{1-r} \\
f(r, 1) & =\int \frac{d r}{1-r}=-\ln (1-r)+f(0,1)
\end{aligned}
$$

By inspection, $f(0,1)=0$, so $f\left(\frac{1}{2}, 1\right)=\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^{n}}=\ln (2)$. It is easy to compute the desired sum in terms of $f\left(\frac{1}{2}, 1\right)$, and we find $\sum_{n=1}^{\infty} \frac{1}{2^{n}(n+4)}=16 \ln (2)-\frac{131}{12}$. Hence, our final answer is $\frac{137}{24}-8 \ln (2)$.