# Use the formula for the sum of a geometric series to find the sum, or state that the series diverges. sum_{n=0}^inftyfrac{3(-2)^n-5^n}{8^n}

Question
Series
Use the formula for the sum of a geometric series to find the sum, or state that the series diverges.
$$\sum_{n=0}^\infty\frac{3(-2)^n-5^n}{8^n}$$

2021-02-01
Given:
$$\sum_{n=0}^\infty\frac{3(-2)^n-5^n}{8^n}$$
To find:
Using the sum geometric series to find the sum of given series.
Consider,
$$\sum_{n=0}^\infty\frac{3(-2)^n-5^n}{8^n}$$
It can be written as,
$$\sum_{n=0}^\infty\frac{3(-2)^n-5^n}{8^n}=\sum_{n=0}^\infty\left[\frac{3(-2)^n}{8^n}-\frac{5^n}{8^n}\right]$$
Apply the sum rule: $$\sum a_n+b_n=\sum a_n+\sum b_n$$
$$=\sum_{n=0}^\infty\frac{3(-2)^n}{8^n}-\sum_{n=0}^\infty\frac{5^n}{8^n}$$
First to find the value of $$\sum_{n=0}^\infty\frac{3(-2)^n}{8^n}$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3(-2)^n}{8^n}=3\sum_{n=0}^\infty\frac{(-2)^n}{8^n}$$
Applying the property: $$\frac{a^n}{b^n}=(\frac{a}{b})^n$$
$$=3\sum_{n=0}^\infty(\frac{-2}{8})^n$$
Using the geometric series:
$$\sum_{n=0}^\infty(\frac{-2}{8})^n=\frac{1}{1-(-\frac{2}{8})}$$
$$=\frac{1}{(1+\frac{2}{8})}$$
$$=\frac{1}{(\frac{8+2}{8})}$$
$$=\frac{1}{(\frac{10}{8})}$$
$$=\frac{8}{10}$$
$$\sum_{n=0}^\infty(\frac{-2}{8})^n=\frac{4}{5}$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3(-2)^n}{8^n}=3\times\frac{4}{5}=\frac{12}{5}$$
Now to find the value of $$\Rightarrow\sum_{n=0}^\infty\frac{5^n}{8^n}$$
$$\Rightarrow\sum_{n=0}^\infty\frac{5^n}{8^n}=(\frac{5}{8})^n$$
$$=\frac{1}{(1-\frac{5}{8})}$$
$$=\frac{1}{(\frac{8-5}{8})}$$
$$=\frac{1}{(\frac{3}{8})}$$
$$=\frac{8}{3}$$
$$\sum_{n=0}^\infty\frac{5^n}{8^n}=\frac{8}{3}$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3(-2)^n-5^n}{8^n}=\frac{12}{5}-\frac{8}{3}$$
$$=\frac{36-40}{15}$$
$$=\frac{-4}{15}$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3(-2)^n-5^n}{8^n}=-\frac{4}{15}$$

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