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

The sum from n=1 to infinity of (5/7)^n is 5/2

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

n = 1 \sum \infty (\frac{5}{7})^{n}

\frac{5}{2}

Solution steps

n = 1 \sum \infty (\frac{5}{7})^{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}{7})^{n} - (\frac{5}{7})^{0}

n = 0 \sum \infty (\frac{5}{7})^{n} = \frac{7}{2}

= \frac{7}{2} - (\frac{5}{7})^{0}

Simplify

\frac{7}{2} - (\frac{5}{7})^{0} : \frac{5}{2}

= \frac{5}{2}

n = 3 \sum \infty \frac{1}{2 ^{n} - 5}

n = 1 \sum \infty (7 + \frac{4}{n})^{- n}

n = 0 \sum \infty \frac{n}{n + 3}

n = 0 \sum \infty 5 + 2 (- 1)^{n}

n = 0 \sum \infty 14 (\frac{2}{3})^{n}

What is the sum from n=1 to infinity of (5/7)^n ?
The sum from n=1 to infinity of (5/7)^n is 5/2