What is the sum from n=1 to infinity of (6/9)^{n-1} ?

The sum from n=1 to infinity of (6/9)^{n-1} is converges

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sum from n=1 to infinity of (6/9)^{n-1}

n = 1 \sum \infty (\frac{6}{9})^{n - 1}

converges

Solution steps

n = 1 \sum \infty (\frac{6}{9})^{n - 1}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty 3 (1 - n)^{- 2}

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

n = 0 \sum \infty (\frac{4}{11})^{n}

j = 1 \sum \infty 2 (- \frac{3}{5})^{j}

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

What is the sum from n=1 to infinity of (6/9)^{n-1} ?
The sum from n=1 to infinity of (6/9)^{n-1} is converges