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

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

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

n = 1 \sum \infty (\frac{99}{100})^{n}

99

Solution steps

n = 1 \sum \infty (\frac{99}{100})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

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

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

= 100 - (\frac{99}{100})^{0}

Simplify

= 99

n = 1 \sum \infty \frac{7}{n ^{3}}

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

n = 0 \sum \infty (- \frac{1}{9})^{n}

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

n = 0 \sum \infty \frac{4}{1 + 3 n ^{3}}

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