Find the sum of the series: sum_{n=0}^infty((frac{5}{2^n})-(frac{1}{3^n}))

Question
Series
asked 2021-01-19
Find the sum of the series:
\(\sum_{n=0}^\infty((\frac{5}{2^n})-(\frac{1}{3^n}))\)

Answers (1)

2021-01-20
The given series is
\(\sum_{n=0}^\infty((\frac{5}{2^n})-(\frac{1}{3^n}))\)
Rewrite the given series as shown below:
\(\sum_{n=0}^\infty((\frac{5}{2^n})-(\frac{1}{3^n}))=\sum_{n=0}^\infty(\frac{5}{2^n})-\sum_{n=0}^\infty(\frac{1}{3^n})\)
From above it can be observed that given series is the sum of two geometric series. Hence, the sum of series is obtained as,
\(\sum_{n=0}^\infty((\frac{5}{2^n})-(\frac{1}{3^n}))=\sum_{n=0}^\infty(\frac{5}{2^n})-\sum_{n=0}^\infty(\frac{1}{3^n})\)
\(=5\sum_{n=0}^\infty(\frac{1}{2^n})-\sum_{n=0}^\infty(\frac{1}{3^n})\)
\(=5(\frac{\frac12}{1-\frac12})-(\frac{\frac13}{1-\frac13})\)
\(=5(\frac{\frac12}{\frac12})-(\frac{\frac13}{\frac23})\)
\(=5(1)-(\frac12)\)
\(=\frac92\)
0

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