What is the sum from n=1 to infinity of 5^{-n} ?

The sum from n=1 to infinity of 5^{-n} is 1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=1 to infinity of 5^{-n}

n = 1 \sum \infty 5^{- n}

\frac{1}{4}

Solution steps

n = 1 \sum \infty 5^{- n}

Simplify

5^{- n} : \frac{1}{5 ^{n}}

= n = 1 \sum \infty \frac{1}{5 ^{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}{5 ^{n}} - \frac{1}{5 ^{0}}

n = 0 \sum \infty \frac{1}{5 ^{n}} = \frac{5}{4}

= \frac{5}{4} - \frac{1}{5 ^{0}}

Simplify

\frac{5}{4} - \frac{1}{5 ^{0}} : \frac{1}{4}

= \frac{1}{4}

\frac{\partial}{\partial x} (arcsin (\frac{l}{x}))

n \to \infty lim (\frac{n}{2})

\frac{d}{d x} (\frac{x + 9}{x})

0.09 e^{x} x

(2, - 6), (- 6, - 10)

What is the sum from n=1 to infinity of 5^{-n} ?
The sum from n=1 to infinity of 5^{-n} is 1/4