What is the sum from n=0 to infinity of (2/7)^n ?

The sum from n=0 to infinity of (2/7)^n is 7/5

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sum from n=0 to infinity of (2/7)^n

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

\frac{7}{5}

Solution steps

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

Apply Series Geometric Test:

\frac{7}{5}

= \frac{7}{5}

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y^{^{''}} + 256 y = 64 cos (7 t)

\int_{0}^{π} sin^{2} (t) d t

\frac{d}{d x} (\frac{4}{x - 2})

\frac{d}{d x} (ln (3 + 2 x))

What is the sum from n=0 to infinity of (2/7)^n ?
The sum from n=0 to infinity of (2/7)^n is 7/5