What is the sum from n=1 to infinity of 0.45^n ?

The sum from n=1 to infinity of 0.45^n is 0.81818…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=1 to infinity of 0.45^n

n = 1 \sum \infty 0.4 5^{n}

0.81818 \dots

Solution steps

n = 1 \sum \infty 0.4 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 0.4 5^{n} - 0.4 5^{0}

n = 0 \sum \infty 0.4 5^{n} = 1.81818 \dots

= 1.81818 \dots - 0.4 5^{0}

Simplify

= 0.81818 \dots

\frac{\partial}{\partial y} (x y^{2} - x^{3} y)

\int 36 - t^{2} d t

d er i v a t i v e f (x) = 2 x^{9} e x

\int x e^{- 4 x^{2}} d x

\int e^{2 x} d x

What is the sum from n=1 to infinity of 0.45^n ?
The sum from n=1 to infinity of 0.45^n is 0.81818…