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

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

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

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

\frac{1}{5}

Solution steps

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

Simplify

= 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}

n = 1 \sum \infty n^{n} (x - 3)^{n}

\int_{0}^{\frac{π}{2}} \frac{cos ( θ )}{1 + sin ( θ )} d θ

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

\frac{d y}{d x} = e^{- (x + y)}

\int e^{- x + 2} d x

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