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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 3 converges

= converges

n = 1 \sum \infty \frac{1}{9 + e ^{- n}}

n = 2 \sum \infty (- 0.5)^{n}

n = 0 \sum \infty \frac{1}{4 + n}

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

n = 0 \sum \infty \frac{4}{n + 3}

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