What is the sum from n=2 to infinity of (3/7)^n ?

The sum from n=2 to infinity of (3/7)^n is 9/28

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sum from n=2 to infinity of (3/7)^n

n = 2 \sum \infty (\frac{3}{7})^{n}

\frac{9}{28}

Solution steps

n = 2 \sum \infty (\frac{3}{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{3}{7})^{n} - (\frac{3}{7})^{0} - (\frac{3}{7})^{1}

n = 0 \sum \infty (\frac{3}{7})^{n} = \frac{7}{4}

= \frac{7}{4} - (\frac{3}{7})^{0} - (\frac{3}{7})^{1}

Simplify

\frac{7}{4} - (\frac{3}{7})^{0} - (\frac{3}{7})^{1} : \frac{9}{28}

= \frac{9}{28}

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

k = 0 \sum \infty \frac{2 ^{k}}{k !}

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

n = 0 \sum \infty \frac{2 ^{2 n}}{n !}

n = 1 \sum \infty e^{0 n}

What is the sum from n=2 to infinity of (3/7)^n ?
The sum from n=2 to infinity of (3/7)^n is 9/28