What is the sum from k=1 to infinity of ke^{-7k} ?

The sum from k=1 to infinity of ke^{-7k} is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from k=1 to infinity of ke^{-7k}

k = 1 \sum \infty k e^{- 7 k}

converges

Solution steps

k = 1 \sum \infty k e^{- 7 k}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty 2 (\frac{3}{2})^{n}

k = 0 \sum \infty (\frac{2}{3})^{k}

n = 1 \sum \infty \frac{1 ^{n}}{n ^{2}}

k = 1 \sum \infty \frac{1}{k ^{2} + 5 k + 6}

n = 1 \sum \infty 3 (0.1)^{n - 1}

What is the sum from k=1 to infinity of ke^{-7k} ?
The sum from k=1 to infinity of ke^{-7k} is converges