What is the sum from n=1 to infinity of (5/12)^n ?

The sum from n=1 to infinity of (5/12)^n is 5/7

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sum from n=1 to infinity of (5/12)^n

n = 1 \sum \infty (\frac{5}{12})^{n}

\frac{5}{7}

Solution steps

n = 1 \sum \infty (\frac{5}{12})^{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{5}{12})^{n} - (\frac{5}{12})^{0}

n = 0 \sum \infty (\frac{5}{12})^{n} = \frac{12}{7}

= \frac{12}{7} - (\frac{5}{12})^{0}

Simplify

\frac{12}{7} - (\frac{5}{12})^{0} : \frac{5}{7}

= \frac{5}{7}

n = 1 \sum \infty (- 2)^{- n}

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

n = 5 \sum \infty \frac{1}{( n - 4 ) ^{3}}

k = 1 \sum \infty \frac{1}{5 ^{k} - 2 ^{k}}

k = 5 \sum \infty \frac{6}{k ln ^{2} ( k )}

What is the sum from n=1 to infinity of (5/12)^n ?
The sum from n=1 to infinity of (5/12)^n is 5/7