What is the sum from n=1 to infinity of 9/(100^n) ?

The sum from n=1 to infinity of 9/(100^n) is 1/11

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sum from n=1 to infinity of 9/(100^n)

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

\frac{1}{11}

Solution steps

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

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

= 9 (n = 0 \sum \infty \frac{1}{10 0 ^{n}} - \frac{1}{10 0 ^{0}})

n = 0 \sum \infty \frac{1}{10 0 ^{n}} = \frac{100}{99}

= 9 (\frac{100}{99} - \frac{1}{10 0 ^{0}})

Simplify

9 (\frac{100}{99} - \frac{1}{10 0 ^{0}}) : \frac{1}{11}

= \frac{1}{11}

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What is the sum from n=1 to infinity of 9/(100^n) ?
The sum from n=1 to infinity of 9/(100^n) is 1/11