What is the sum from n=1 to infinity of (0.3)^n ?

The sum from n=1 to infinity of (0.3)^n is 0.42857…

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sum from n=1 to infinity of (0.3)^n

n = 1 \sum \infty (0.3)^{n}

0.42857 \dots

Solution steps

n = 1 \sum \infty 0. 3^{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. 3^{n} - 0. 3^{0}

n = 0 \sum \infty 0. 3^{n} = 1.42857 \dots

= 1.42857 \dots - 0. 3^{0}

Simplify

= 0.42857 \dots

k = 1 \sum \infty \frac{1}{k ^{3} + k ^{2}}

n = 0 \sum \infty \frac{5}{1 + 2 n ^{4}}

n = 2 \sum \infty (- 0.05)^{n}

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

n = 1 \sum \infty n! e^{- 3 n}

What is the sum from n=1 to infinity of (0.3)^n ?
The sum from n=1 to infinity of (0.3)^n is 0.42857…