What is the sum from n=0 to infinity of (1/100)^n ?

The sum from n=0 to infinity of (1/100)^n is 100/99

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sum from n=0 to infinity of (1/100)^n

n = 0 \sum \infty (\frac{1}{100})^{n}

\frac{100}{99}

Solution steps

n = 0 \sum \infty (\frac{1}{100})^{n}

Apply Series Geometric Test:

\frac{100}{99}

= \frac{100}{99}

n = 0 \sum \infty \frac{9}{2 ^{n}}

n = 1 \sum \infty \frac{n ^{4}}{8 ^{n}}

n = 1 \sum \infty \frac{n}{n ^{2} + 16}

k = 1 \sum \infty (\frac{π}{e})^{- k}

n = 1 \sum \infty \frac{n}{cos ( n )}

What is the sum from n=0 to infinity of (1/100)^n ?
The sum from n=0 to infinity of (1/100)^n is 100/99