What is the sum from n=1 to infinity of (4/10)^n ?

The sum from n=1 to infinity of (4/10)^n is 2/3

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sum from n=1 to infinity of (4/10)^n

n = 1 \sum \infty (\frac{4}{10})^{n}

\frac{2}{3}

Solution steps

n = 1 \sum \infty (\frac{4}{10})^{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{4}{10})^{n} - (\frac{4}{10})^{0}

n = 0 \sum \infty (\frac{4}{10})^{n} = \frac{5}{3}

= \frac{5}{3} - (\frac{4}{10})^{0}

Simplify

\frac{5}{3} - (\frac{4}{10})^{0} : \frac{2}{3}

= \frac{2}{3}

\int \frac{2 x + 2}{x ^{2} - 4 x + 9} d x

\int - t e^{- s t} d t

\int_{- 1}^{2} ∣ e^{x} - 1 ∣ d x

\frac{2}{s + 1} + \frac{32 s}{s ^{2} + 9}

\frac{d}{d x} (2 x^{- \frac{1}{2}})

What is the sum from n=1 to infinity of (4/10)^n ?
The sum from n=1 to infinity of (4/10)^n is 2/3