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

The sum from n=2 to infinity of (2/7)^n is 4/35

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

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

\frac{4}{35}

Decimal

0.11428 \dots

Solution steps

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

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

= \frac{7}{5} - (\frac{2}{7})^{0} - (\frac{2}{7})^{1}

Simplify

\frac{7}{5} - (\frac{2}{7})^{0} - (\frac{2}{7})^{1} : \frac{4}{35}

= \frac{4}{35}

What is the sum from n=2 to infinity of (2/7)^n ?
The sum from n=2 to infinity of (2/7)^n is 4/35