What is the sum from n=0 to infinity of (1/20)^n ?

The sum from n=0 to infinity of (1/20)^n is 20/19

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sum from n=0 to infinity of (1/20)^n

n = 0 \sum \infty (\frac{1}{20})^{n}

\frac{20}{19}

Solution steps

n = 0 \sum \infty (\frac{1}{20})^{n}

Apply Series Geometric Test:

\frac{20}{19}

= \frac{20}{19}

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

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

n = 1 \sum \infty \frac{1}{( 4 n + 5 ) ^{3}}

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

n = 0 \sum \infty \frac{4}{2 ^{2 n - 2}}

What is the sum from n=0 to infinity of (1/20)^n ?
The sum from n=0 to infinity of (1/20)^n is 20/19