What is the sum from k=0 to infinity of (1/19)^k ?

The sum from k=0 to infinity of (1/19)^k is 19/18

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sum from k=0 to infinity of (1/19)^k

k = 0 \sum \infty (\frac{1}{19})^{k}

\frac{19}{18}

Solution steps

k = 0 \sum \infty (\frac{1}{19})^{k}

Apply Series Geometric Test:

\frac{19}{18}

= \frac{19}{18}

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

n = 0 \sum \infty 6 (\frac{8}{9})^{n}

n = 0 \sum \infty 0.5

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

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

What is the sum from k=0 to infinity of (1/19)^k ?
The sum from k=0 to infinity of (1/19)^k is 19/18