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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{1 7 ^{n}}{n !}

n = 1 \sum \infty cos (\frac{18}{n})

n = 1 \sum \infty \frac{n ^{5}}{6 ^{n}}

k = 0 \sum \infty (\frac{2}{5})^{k}

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

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