What is the sum from n=2 to infinity of 1/(e^{3n)} ?

The sum from n=2 to infinity of 1/(e^{3n)} is converges

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sum from n=2 to infinity of 1/(e^{3n)}

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

converges

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{3 n}} = e^{- 3 n}

= n = 2 \sum \infty e^{- 3 n}

Apply Series Ratio Test:

converges

= converges

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

n = 1 \sum \infty \frac{1}{4 n ^{2} + 1}

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

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

k = 1 \sum \infty \frac{3}{k !}

What is the sum from n=2 to infinity of 1/(e^{3n)} ?
The sum from n=2 to infinity of 1/(e^{3n)} is converges