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

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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

n = 3 \sum \infty \frac{1}{n ^{\frac{3}{2}}}

n = 0 \sum \infty (\frac{- 7}{8})^{n}

k = 1 \sum \infty 1^{k}

n = 0 \sum \infty 2021

n = 1 \sum \infty (- 1)^{n} \frac{3 n}{7 n ^{2} + 1}

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