Calculate the value of the following fraction, supposing that in the numerator and in the denominator there are infinite terms:
$$\dfrac{-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\ldots}{\dfrac{3}{5}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{45}+\ldots}$$
Development:
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:
$$$\dfrac{-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\ldots}{\dfrac{3}{5}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{45}+\ldots} = \dfrac{-1}{\dfrac{9}{10}}=-\dfrac{10}{9}$$$
Solution:
$$-\dfrac{10}{9}$$