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}

1 + x^{2}

f (x) = 5 x^{2}, at x = - 2

y = t (ln (4 t))^{2}

h \to 0 lim (\frac{2 ( x + h ) - 2 x}{h})

\int \frac{1}{- y} d y

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