What is the sum from n=0 to infinity of (1/16)^n ?

The sum from n=0 to infinity of (1/16)^n is 16/15

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sum from n=0 to infinity of (1/16)^n

n = 0 \sum \infty (\frac{1}{16})^{n}

\frac{16}{15}

Solution steps

n = 0 \sum \infty (\frac{1}{16})^{n}

Apply Series Geometric Test:

\frac{16}{15}

= \frac{16}{15}

n = 1 \sum \infty \frac{1}{n ^{1.03}}

n = 0 \sum \infty sin (3 n)

n = 0 \sum \infty \frac{1}{( 1 + n ) ^{4}}

n = 1 \sum \infty (\frac{4}{9})^{n - 1}

n = 0 \sum \infty (3.5)^{- n}

What is the sum from n=0 to infinity of (1/16)^n ?
The sum from n=0 to infinity of (1/16)^n is 16/15