What is the sum from n=1 to infinity of 3e^{-9n} ?

The sum from n=1 to infinity of 3e^{-9n} 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 3e^{-9n}

n = 1 \sum \infty 3 e^{- 9 n}

converges

Solution steps

n = 1 \sum \infty 3 e^{- 9 n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot n = 1 \sum \infty e^{- 9 n}

Apply Series Ratio Test:

converges

= 3 converges

= converges

f (x) = \frac{1}{x ^{2}}, [2, 6]

(ln (x))^{4}

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

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

cos (t) \cdot cos (π)

What is the sum from n=1 to infinity of 3e^{-9n} ?
The sum from n=1 to infinity of 3e^{-9n} is converges