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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=1 to infinity of 1/(1-e^{-n)}

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

diverges

Solution steps

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

Simplify

e^{- n} : \frac{1}{e ^{n}}

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

Simplify

\frac{1}{1 - \frac{1}{e ^{n}}} : \frac{e ^{n}}{e ^{n} - 1}

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

Apply Series Divergence Test:

diverges

= diverges

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