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

\frac{\partial}{\partial x} (- 3 x y^{2} - 2 y cos (x))

\frac{d}{d x} (x y^{2} z^{3})

7 x^{2} y^{^{'}} = y^{^{'}} + 7 x e^{- y}

y^{^{'}} + \frac{2}{x} \cdot y = 0

x \to - 4 - lim (\frac{3}{x ^{2} - 16})

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…