What is the sum from n=1 to infinity of 1/2 (0.3)^n ?

The sum from n=1 to infinity of 1/2 (0.3)^n is 0.21428…

sum from n=1 to infinity of 1/2 (0.3)^n

n = 1 \sum \infty \frac{1}{2} (0.3)^{n}

0.21428 \dots

Solution steps

n = 1 \sum \infty \frac{1}{2} \cdot 0. 3^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= \frac{1}{2} \cdot 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}

= \frac{1}{2} (n = 0 \sum \infty 0. 3^{n} - 0. 3^{0})

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

= \frac{1}{2} (1.42857 \dots - 0. 3^{0})

Simplify

\frac{1}{2} (1.42857 \dots - 0. 3^{0}) : 0.21428 \dots

= 0.21428 \dots

\int \frac{1 - x ^{2}}{1 - 2 x} d x

\int_{\frac{1}{2}}^{3} 16 x ln (2 x) d x

\frac{\partial}{\partial x} (ln (\frac{r}{ro}))

\frac{d}{d x} (e^{j x})

y^{^{'}} = y^{2} - 9

What is the sum from n=1 to infinity of 1/2 (0.3)^n ?
The sum from n=1 to infinity of 1/2 (0.3)^n is 0.21428…