Evaluating $\frac{13}{1.2 .3 .2}+\frac{26}{2.3 .4 .4}+\frac{43}{3.4 .5 .8}+\frac{64}{4.5 .6 .16}+\cdots$ $$\frac{13}{1.2 .3 .2}+\frac{26}{2.3 .4 .4}+\frac{43}{3.4 .5 .8}+\frac{64}{4.5 .6 .16}+\cdots$$
I can reduce it to the general term,
$$\sum_{r=1}^\infty \frac{2r^2 + 7r +4}{r(r+1)(r+2)2^r}$$
I don't know how to go about this any further though. I also ran this in python and the sum is exceeding $1.5$ for $10,000$ terms, which is weird since it should converge to $1.5$, so it makes me doubt if the general term I've written is correct.
 A: Consider $$f(x)=\sum_{r=1}^\infty \frac{2r^2 + 7r +4}{r(r+1)(r+2)}x^r=\sum_{r=1}^\infty \left(\frac{1}{r+1}-\frac{1}{r+2}+\frac{2}{r} \right)x^r$$
$$f(x)=\frac{\log (1-x)}{x^2}+\frac{1}{x}-\frac{\log (1-x)}{x}-2 \log
   (1-x)-\frac{1}{2}$$
$$f\left(\frac{1}{2}\right)=\frac 3 2$$
A: Using telescoping:
$$\begin{align}
\sum_{n=1}^\infty\frac{2n^2+7n+4}{n(n+1)(n+2)2^n}&=
\sum_{n=1}^\infty\left[\frac2n+\frac1{n+1}-\frac1{n+2}\right]\frac1{2^n}\\
&=\sum_{n=1}^\infty\left[\left(\frac1n+\frac1{n+1}\right)
-\frac12\left(\frac1{n+1}+\frac1{n+2}\right)\right]\frac1{2^{n-1}}\\
&=1+\frac12=\frac32.
\end{align}$$
A: Here is a longer and more painstaking way that uses calculus:
For $|x|\lt 1$, $$\sum_1^{\infty}(2r^2 +7r+4) x^{r-1}\\ = 2\sum_1^{\infty} r^2 x^{r-1} +7\sum_1^{\infty} rx^{r-1} +4\sum_1^{\infty}x^{r-1}  \\ = 2\frac{d}{dx} \left( x \frac{dS}{dx}\right ) +7\frac{dS}{dx} +4S \\ = \frac{-2(1+x)}{(1-x)^3} +\frac{7}{(1-x)^2} +\frac{4}{1-x} $$
where $S=\sum_0^{\infty} x^r =\frac{1}{1-x}$. Integrating both sides thrice, dividing by $x^2$ and setting $x=\frac 12$ will yield the exact sum you mention.
