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

The sum from n=3 to infinity of (1/4)^n is 1/48

Popular >

sum from n=3 to infinity of (1/4)^n

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

\frac{1}{48}

Solution steps

n = 3 \sum \infty (\frac{1}{4})^{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{1}{4})^{n} - (\frac{1}{4})^{0} - (\frac{1}{4})^{1} - (\frac{1}{4})^{2}

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

= \frac{4}{3} - (\frac{1}{4})^{0} - (\frac{1}{4})^{1} - (\frac{1}{4})^{2}

Simplify

\frac{4}{3} - (\frac{1}{4})^{0} - (\frac{1}{4})^{1} - (\frac{1}{4})^{2} : \frac{1}{48}

= \frac{1}{48}

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

n = 0 \sum \infty 3 (\frac{2}{6})^{n + 2}

k = 1 \sum \infty \frac{k !}{10 4 ^{k}}

n = 1 \sum \infty n arctan (\frac{1}{n})

n = 2 \sum \infty (\frac{3}{7})^{4 n}

What is the sum from n=3 to infinity of (1/4)^n ?
The sum from n=3 to infinity of (1/4)^n is 1/48