What is the sum from n=0 to infinity of 9/(100^n) ?

The sum from n=0 to infinity of 9/(100^n) is 100/11

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sum from n=0 to infinity of 9/(100^n)

n = 0 \sum \infty \frac{9}{10 0 ^{n}}

\frac{100}{11}

Solution steps

n = 0 \sum \infty \frac{9}{10 0 ^{n}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 9 \cdot n = 0 \sum \infty \frac{1}{10 0 ^{n}}

Apply Series Geometric Test:

\frac{100}{99}

= 9 \cdot \frac{100}{99}

Simplify

9 \cdot \frac{100}{99} : \frac{100}{11}

= \frac{100}{11}

\frac{d y}{d x} = \frac{y}{x} ln (\frac{y}{x})

\frac{\partial ^{2}}{\partial r \partial θ} (r (θ) θ)

\frac{d}{d x} ((x + 1) e^{x})

\int x^{11} sin (x^{12}) d x

x + 4 ln (x + 10) - ln (2 x + 1)

What is the sum from n=0 to infinity of 9/(100^n) ?
The sum from n=0 to infinity of 9/(100^n) is 100/11