What is the sum from n=1 to infinity of (1/6)^n ?

The sum from n=1 to infinity of (1/6)^n is 1/5

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sum from n=1 to infinity of (1/6)^n

n = 1 \sum \infty (\frac{1}{6})^{n}

\frac{1}{5}

Solution steps

n = 1 \sum \infty (\frac{1}{6})^{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{1}{6})^{n} - (\frac{1}{6})^{0}

n = 0 \sum \infty (\frac{1}{6})^{n} = \frac{6}{5}

= \frac{6}{5} - (\frac{1}{6})^{0}

Simplify

\frac{6}{5} - (\frac{1}{6})^{0} : \frac{1}{5}

= \frac{1}{5}

x^{2} + 8 x - 7

\frac{9 x}{2 ^{x}}

y = 4 x^{2} + 2

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

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

What is the sum from n=1 to infinity of (1/6)^n ?
The sum from n=1 to infinity of (1/6)^n is 1/5