What is the sum from n=0 to infinity of n/(e^{7n)} ?

The sum from n=0 to infinity of n/(e^{7n)} is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=0 to infinity of n/(e^{7n)}

n = 0 \sum \infty \frac{n}{e ^{7 n}}

converges

Solution steps

n = 0 \sum \infty \frac{n}{e ^{7 n}}

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{n}{e ^{7 n}} = (n) e^{- 7 n}

= n = 0 \sum \infty n e^{- 7 n}

Apply Series Ratio Test:

converges

= converges

What is the sum from n=0 to infinity of n/(e^{7n)} ?
The sum from n=0 to infinity of n/(e^{7n)} is converges