What is the sum from n=1 to infinity of 1/(12^n) ?

The sum from n=1 to infinity of 1/(12^n) is 1/11

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sum from n=1 to infinity of 1/(12^n)

n = 1 \sum \infty \frac{1}{1 2 ^{n}}

\frac{1}{11}

Solution steps

n = 1 \sum \infty \frac{1}{1 2 ^{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}{1 2 ^{n}} - \frac{1}{1 2 ^{0}}

n = 0 \sum \infty \frac{1}{1 2 ^{n}} = \frac{12}{11}

= \frac{12}{11} - \frac{1}{1 2 ^{0}}

Simplify

\frac{12}{11} - \frac{1}{1 2 ^{0}} : \frac{1}{11}

= \frac{1}{11}

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

n = 1 \sum \infty (\frac{8 π}{9})^{n}

n = 2 \sum \infty \frac{n !}{6 ^{n}}

k = 1 \sum \infty \frac{6}{k ( k + 2 )}

k = 1 \sum \infty \frac{3}{k}

What is the sum from n=1 to infinity of 1/(12^n) ?
The sum from n=1 to infinity of 1/(12^n) is 1/11