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

The sum from k=1 to infinity of (1/4)^k is 1/3

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

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

\frac{1}{3}

Solution steps

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

k = 0 \sum \infty (\frac{1}{4})^{k} = \frac{4}{3}

= \frac{4}{3} - (\frac{1}{4})^{0}

Simplify

\frac{4}{3} - (\frac{1}{4})^{0} : \frac{1}{3}

= \frac{1}{3}

n = 1 \sum \infty \frac{8}{e ^{n} - 1}

n = 0 \sum \infty \frac{1}{2 n ^{3}}

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

n = 0 \sum \infty \frac{n + 1}{3 n + 1}

n = 0 \sum \infty e^{- 14 n}

What is the sum from k=1 to infinity of (1/4)^k ?
The sum from k=1 to infinity of (1/4)^k is 1/3