What is the sum from n=0 to infinity of 3/(5^{n-1)} ?

The sum from n=0 to infinity of 3/(5^{n-1)} is converges

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sum from n=0 to infinity of 3/(5^{n-1)}

n = 0 \sum \infty \frac{3}{5 ^{n - 1}}

converges

Solution steps

n = 0 \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 = 0 \sum \infty \frac{1}{5 ^{n - 1}}

Apply Series Ratio Test:

converges

= 3 converges

= converges

n = 0 \sum \infty n - \frac{π}{2}

x = 1 \sum \infty \frac{1}{x ^{2}}

k = 1 \sum \infty (- \frac{1}{4})^{k}

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

n = 0 \sum \infty \frac{7}{n ^{2} + 9}

What is the sum from n=0 to infinity of 3/(5^{n-1)} ?
The sum from n=0 to infinity of 3/(5^{n-1)} is converges