What is the sum from n=1 to infinity of 7e^{-2n} ?

The sum from n=1 to infinity of 7e^{-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=1 to infinity of 7e^{-2n}

n = 1 \sum \infty 7 e^{- 2 n}

converges

Solution steps

n = 1 \sum \infty 7 e^{- 2 n}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty e^{- 2 n}

Apply Series Ratio Test:

converges

= 7 converges

= converges

\int (x^{2} + y^{2}) d x

\int 4 - x^{2} d x

0.81 t^{2}

\int x e^{11 x} d x

f (x) = x^{- 11}

What is the sum from n=1 to infinity of 7e^{-2n} ?
The sum from n=1 to infinity of 7e^{-2n} is converges