What is the sum from n=0 to infinity of 1-e^{-n} ?

The sum from n=0 to infinity of 1-e^{-n} is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of 1-e^{-n}

n = 0 \sum \infty 1 - e^{- n}

diverges

Solution steps

n = 0 \sum \infty 1 - e^{- n}

Apply Series Divergence Test:

diverges

= diverges

k = 1 \sum \infty \frac{k}{k ^{2} + 4}

n = 2 \sum \infty \frac{1}{( n - 1 ) ^{4}}

n = 0 \sum \infty 5 (- \frac{3}{2})^{n}

n = 15 \sum \infty (\frac{2}{3})^{n}

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

What is the sum from n=0 to infinity of 1-e^{-n} ?
The sum from n=0 to infinity of 1-e^{-n} is diverges