What is the sum from n=0 to infinity of (-2/5)^n ?

The sum from n=0 to infinity of (-2/5)^n is 5/7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of (-2/5)^n

n = 0 \sum \infty (- \frac{2}{5})^{n}

\frac{5}{7}

Solution steps

n = 0 \sum \infty (- \frac{2}{5})^{n}

Apply Series Geometric Test:

\frac{5}{7}

= \frac{5}{7}

\frac{8}{s ^{3}}

\frac{d}{d x} (- \frac{4}{x})

x \to \frac{π}{2} lim (2 tan (x))

\int \frac{8 - 128 x}{25 - 64 x ^{2}} d x

f (x) = x^{- 6}

What is the sum from n=0 to infinity of (-2/5)^n ?
The sum from n=0 to infinity of (-2/5)^n is 5/7