Evaluate an Infinite Sum

$\begin{aligned} S & = \sum_{\blue{n=1}}^\infty \frac {4n+5}{3^{n-1}}\left(\frac 12\right)^{n+2} \\ & = \frac 12 \sum_{\blue{n=1}}^\infty \frac n{6^{n-1}} + \frac 58 \sum_{\red{n=0}}^\infty \left(\frac 16\right)^n \\ & = \frac 12 \sum_{n=1}^\infty \frac d{dx}x^n\ \bigg|_{x = \frac 16} + \frac 58 \left(\frac 1{1-\frac 16}\right) \\ & = \frac 12 \cdot \frac d{dx} \sum_{n=1}^\infty x^n\ \bigg|_{x = \frac 16} + \frac 34 \\ & = \frac 12 \cdot \frac d{dx} \left(\frac x{1-x}\right) \bigg|_{x = \frac 16} + \frac 34 \\ & = \frac 1{2(1-x)^2} \bigg|_{x = \frac 16} + \frac 34 \\ & = \frac {18}{25} + \frac 34 = \frac {147}{100} = \boxed{1.47} \end{aligned}$

$\sum_{n=1}^\infty \frac{4n+5}{3^{n-1}}(\frac{1}{2})^{n+2}$

$=\frac{3}{4}\sum_{n=1}^\infty \frac{4n+5}{3^{n}}(\frac{1}{2})^{n}$

$=\frac{3}{4}\sum_{n=1}^\infty \frac{4n}{3^{n}}(\frac{1}{2})^{n} + \frac{3}{4}\sum_{n=1}^\infty \frac{5}{3^{n}}(\frac{1}{2})^{n}$

$=3\sum_{n=1}^\infty \frac{n}{6^{n}} + \frac{15}{4}\sum_{n=1}^\infty (\frac{1}{6})^{n}$

$=3\sum_{n=1}^\infty \frac{n}{6^{n}} + \frac{15}{4}(\frac{\frac{1}{6}}{1-\frac{1}{6}})$

we can notice that

$\frac{d}{dx} \sum_{n=0}^\infty x^{n} = \frac{d}{dx} \frac{1}{1-x}$

$\sum_{n=1}^\infty nx^{n-1} = \frac{1}{(1-x)^{2}}$

$\sum_{n=1}^\infty nx^{n} = \frac{x}{(1-x)^{2}}$

$\sum_{n=1}^\infty \frac{n}{\frac{1}{x}^{n}} = \frac{x}{(1-x)^{2}}$

so, with x = 1/6

$3\sum_{n=1}^\infty \frac{n}{6^{n}} + \frac{15}{4}(\frac{\frac{1}{6}}{1-\frac{1}{6}})$

$=3(\frac{6}{(6-1)^{2}}) + \frac{3}{4}$

$=\frac{18}{25} + \frac{3}{4}$

$=\frac{147}{100}$

$=\boxed{1.47}$