What is the sum from n=1 to infinity of (10/11)^n ?

The sum from n=1 to infinity of (10/11)^n is 10

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sum from n=1 to infinity of (10/11)^n

n = 1 \sum \infty (\frac{10}{11})^{n}

10

Solution steps

n = 1 \sum \infty (\frac{10}{11})^{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{10}{11})^{n} - (\frac{10}{11})^{0}

n = 0 \sum \infty (\frac{10}{11})^{n} = 11

= 11 - (\frac{10}{11})^{0}

Simplify

= 10

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What is the sum from n=1 to infinity of (10/11)^n ?
The sum from n=1 to infinity of (10/11)^n is 10