What is the sum from n=1 to infinity of (4/9)^n ?

The sum from n=1 to infinity of (4/9)^n is 4/5

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sum from n=1 to infinity of (4/9)^n

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

\frac{4}{5}

Solution steps

n = 1 \sum \infty (\frac{4}{9})^{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{4}{9})^{n} - (\frac{4}{9})^{0}

n = 0 \sum \infty (\frac{4}{9})^{n} = \frac{9}{5}

= \frac{9}{5} - (\frac{4}{9})^{0}

Simplify

\frac{9}{5} - (\frac{4}{9})^{0} : \frac{4}{5}

= \frac{4}{5}

n = 0 \sum \infty n^{- n}

n = 1 \sum \infty (\frac{π}{e})^{n}

n = 1 \sum \infty \frac{1}{- n}

n = 1 \sum \infty \frac{n + 5}{n + 4}

n = 2 \sum \infty ln (\frac{n + 1}{n})

What is the sum from n=1 to infinity of (4/9)^n ?
The sum from n=1 to infinity of (4/9)^n is 4/5