What is the sum from n=1 to infinity of 0.45^n ?

The sum from n=1 to infinity of 0.45^n is 0.81818…

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sum from n=1 to infinity of 0.45^n

n = 1 \sum \infty 0.4 5^{n}

0.81818 \dots

Solution steps

n = 1 \sum \infty 0.4 5^{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 0.4 5^{n} - 0.4 5^{0}

n = 0 \sum \infty 0.4 5^{n} = 1.81818 \dots

= 1.81818 \dots - 0.4 5^{0}

Simplify

= 0.81818 \dots

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What is the sum from n=1 to infinity of 0.45^n ?
The sum from n=1 to infinity of 0.45^n is 0.81818…