What is the sum from n=0 to infinity of e^{-0.2n} ?

The sum from n=0 to infinity of e^{-0.2n} is converges

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 e^{-0.2n}

n = 0 \sum \infty e^{- 0.2 n}

converges

Solution steps

n = 0 \sum \infty e^{- 0.2 n}

n \to \infty lim (n (\frac{e ^{- 0.2 n}}{e ^{- 0.2 (n + 1)}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

n = 1 \sum \infty \frac{4}{n - 1}

n = 0 \sum \infty 2 (\frac{1}{5})^{n + 1}

k = 1 \sum \infty \frac{1}{k ( k + 2 )}

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

n = 0 \sum \infty (0.03)^{n}

What is the sum from n=0 to infinity of e^{-0.2n} ?
The sum from n=0 to infinity of e^{-0.2n} is converges