What is the sum from k=1 to infinity of (2/5)^k ?

The sum from k=1 to infinity of (2/5)^k is 2/3

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sum from k=1 to infinity of (2/5)^k

k = 1 \sum \infty (\frac{2}{5})^{k}

\frac{2}{3}

Solution steps

k = 1 \sum \infty (\frac{2}{5})^{k}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= k = 0 \sum \infty (\frac{2}{5})^{k} - (\frac{2}{5})^{0}

k = 0 \sum \infty (\frac{2}{5})^{k} = \frac{5}{3}

= \frac{5}{3} - (\frac{2}{5})^{0}

Simplify

\frac{5}{3} - (\frac{2}{5})^{0} : \frac{2}{3}

= \frac{2}{3}

n = 5 \sum \infty \frac{1}{ln ( n - 3 )}

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

n = 0 \sum \infty \frac{5}{8 ^{n}}

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

k = 1 \sum \infty (cos (1))^{k}

What is the sum from k=1 to infinity of (2/5)^k ?
The sum from k=1 to infinity of (2/5)^k is 2/3