What is the sum from n=1 to infinity of (7/10)^n ?

The sum from n=1 to infinity of (7/10)^n is 7/3

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sum from n=1 to infinity of (7/10)^n

n = 1 \sum \infty (\frac{7}{10})^{n}

\frac{7}{3}

Solution steps

n = 1 \sum \infty (\frac{7}{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{7}{10})^{n} - (\frac{7}{10})^{0}

n = 0 \sum \infty (\frac{7}{10})^{n} = \frac{10}{3}

= \frac{10}{3} - (\frac{7}{10})^{0}

Simplify

\frac{10}{3} - (\frac{7}{10})^{0} : \frac{7}{3}

= \frac{7}{3}

\int \frac{1 + e ^{x}}{1 - e ^{x}} d x

h \to 0 lim (\frac{( 7 + h ) - 7}{h})

\int \frac{x + 1}{x + 1} d x

t^{2} (8 e^{2 t} - 6 sin (t))

x \to 0 lim ((1 + 4 x)^{\frac{4}{x}})

What is the sum from n=1 to infinity of (7/10)^n ?
The sum from n=1 to infinity of (7/10)^n is 7/3