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
asked 2021-01-31
Use the formula for the sum of a geometric series to find the sum, or state that the series diverges.
\(\sum_{n=0}^\infty\frac{3(-2)^n-5^n}{8^n}\)

Answers (1)

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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