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

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

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

n = 1 \sum \infty (\frac{π}{6})^{n}

\frac{6}{6 - π} - 1

Decimal

1.09907 \dots

Solution steps

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

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

= \frac{6}{6 - π} - (\frac{π}{6})^{0}

Simplify

= \frac{6}{6 - π} - 1

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 7 + e^{\frac{x}{2}}

\int t (t^{6} + t - 5) d t

x \to \infty lim (3 e^{- 3 x})

d er i v a t i v e f (x) = (x^{3} - 5 x) (5 x^{2} + 3 x)

\frac{\partial}{\partial x} ([A x^{2} + B y^{3}]^{\frac{1}{4}})

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