What is the sum from n=1 to infinity of e^{-3n} ?

The sum from n=1 to infinity of e^{-3n} is converges

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sum from n=1 to infinity of e^{-3n}

n = 1 \sum \infty e^{- 3 n}

converges

Solution steps

n = 1 \sum \infty e^{- 3 n}

Apply Series Ratio Test:

converges

= converges

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

n = 1 \sum \infty 1 - n

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

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

n = 1 \sum \infty \frac{1}{n + 6}

What is the sum from n=1 to infinity of e^{-3n} ?
The sum from n=1 to infinity of e^{-3n} is converges