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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty 8 \cdot (\frac{5}{8})^{n}

n = 0 \sum \infty 5 + \frac{15}{5} + \frac{45}{16}

n = 1 \sum \infty \frac{2}{2 n + 1}

n = 2 \sum \infty (\frac{3}{7})^{n}

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

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