What is the sum from n=1 to infinity of 5/(3^{n-1)} ?

The sum from n=1 to infinity of 5/(3^{n-1)} is converges

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sum from n=1 to infinity of 5/(3^{n-1)}

n = 1 \sum \infty \frac{5}{3 ^{n - 1}}

converges

Solution steps

n = 1 \sum \infty \frac{5}{3 ^{n - 1}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 5 \cdot n = 1 \sum \infty \frac{1}{3 ^{n - 1}}

Apply Series Ratio Test:

converges

= 5 converges

= converges

n = 1 \sum \infty \frac{π}{n ^{2}}

n = 1 \sum \infty (\frac{- 1}{3})^{n}

i = 1 \sum \infty 4 (\frac{2}{5})^{i - 1}

n = 1 \sum \infty \frac{4}{n ^{2} + n ^{3}}

k = 3 \sum \infty \frac{5}{k} - \frac{5}{k + 1}

What is the sum from n=1 to infinity of 5/(3^{n-1)} ?
The sum from n=1 to infinity of 5/(3^{n-1)} is converges