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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{\partial}{\partial x} (y cos (2 x))

\int (tan^{2} (x) + cot^{2} (x) + 4) d x

\frac{x - 9 x x}{x}

y (2 x + x^{2} cot (x)) d x + x^{2} (1 + y) d y = 0

e^{3 x}

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