Problems from Infinite sums of series

Development:

We firstly study the value of the numerator: $-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{16}}-\dots$

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

$$-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{16}}-\dots =\sum _{n\ge 1}-{\displaystyle \frac{1}{2^n}}={\displaystyle \frac{-{\displaystyle \frac{1}{2}}}{1-{\displaystyle \frac{1}{2}}}}=-1$$

Next we look at the value of the denominator: ${\displaystyle \frac{3}{5}}+{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{15}}+{\displaystyle \frac{1}{45}}+\dots$

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

$${\displaystyle \frac{3}{5}}+{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{15}}+{\displaystyle \frac{1}{45}}+\dots =\sum _{n\ge 1}{\displaystyle \frac{1}{5\cdot 3^{n-2}}}={\displaystyle \frac{{\displaystyle \frac{3}{5}}}{1-{\displaystyle \frac{1}{3}}}}={\displaystyle \frac{9}{10}}$$

This way we have:

$${\displaystyle \frac{-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{16}}-\dots }{{\displaystyle \frac{3}{5}}+{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{15}}+{\displaystyle \frac{1}{45}}+\dots }}={\displaystyle \frac{-1}{{\displaystyle \frac{9}{10}}}}=-{\displaystyle \frac{10}{9}}$$

Solution:

$-{\displaystyle \frac{10}{9}}$

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