Infinite sums of series

We firstly study the value of the numerator: $-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\ldots$

This is the sum of a geometric progression of the first term $a_1=-\dfrac{1}{2}$, and ratio $r=\dfrac{1}{2}$, so it is:

$$-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\ldots = \sum_{n\geq 1}-\dfrac{1}{2^n}=\dfrac{-\dfrac{1}{2}}{1-\dfrac{1}{2}}=-1$$

Next we look at the value of the denominator: $$\dfrac{3}{5}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{45}+\ldots$$

It is the sum of a geometric progression, this time the first term is $b_1=\dfrac{3}{5}$ and ratio $r=\dfrac{1}{3}$, so:

$$\dfrac{3}{5}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{45}+\ldots = \sum_{n\geq 1}\dfrac{1}{5\cdot 3^{n-2}}=\dfrac{\dfrac{3}{5}}{1-\dfrac{1}{3}}=\dfrac{9}{10}$$

This way we have:

$$\dfrac{-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\ldots}{\dfrac{3}{5}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{45}+\ldots} = \dfrac{-1}{\dfrac{9}{10}}=-\dfrac{10}{9}$$