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

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

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

n = 0 \sum \infty (\frac{99}{100})^{n}

100

Solution steps

n = 0 \sum \infty (\frac{99}{100})^{n}

Apply Series Geometric Test:

100

= 100

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

n = 1 \sum \infty (\frac{2}{10})^{n}

n = 0 \sum \infty \frac{2}{3 n + 5}

n = 1 \sum \infty \frac{6 n}{e ^{n}}

n = 1 \sum \infty n

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