What is the sum from n=1 to infinity of 3/(9^{n+1)} ?

The sum from n=1 to infinity of 3/(9^{n+1)} is converges

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sum from n=1 to infinity of 3/(9^{n+1)}

n = 1 \sum \infty \frac{3}{9 ^{n + 1}}

converges

Solution steps

n = 1 \sum \infty \frac{3}{9 ^{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}{9 ^{n + 1}}

Apply Series Ratio Test:

converges

= 3 converges

= converges

n = 1 \sum \infty 2 n + 3

n = 1 \sum \infty \frac{1}{8 n - 1}

n = 1 \sum \infty (\frac{1}{n})^{2}

n = 0 \sum \infty cos (8 n)

n = 0 \sum \infty \frac{1}{2 + 3 ^{- n}}

What is the sum from n=1 to infinity of 3/(9^{n+1)} ?
The sum from n=1 to infinity of 3/(9^{n+1)} is converges