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

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{1 + 3 n ^{3}}

n = 1 \sum \infty (\frac{5}{7})^{n}

n = 3 \sum \infty \frac{1}{2 ^{n} - 5}

n = 1 \sum \infty (7 + \frac{4}{n})^{- n}

n = 0 \sum \infty \frac{n}{n + 3}

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