What is the sum from n=2 to infinity of e^{3-6n} ?

The sum from n=2 to infinity of e^{3-6n} is converges

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sum from n=2 to infinity of e^{3-6n}

n = 2 \sum \infty e^{3 - 6 n}

converges

Solution steps

n = 2 \sum \infty e^{3 - 6 n}

Simplify

e^{3 - 6 n} : e^{3} e^{- 6 n}

= n = 2 \sum \infty e^{3} e^{- 6 n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= e^{3} \cdot n = 2 \sum \infty e^{- 6 n}

Apply Series Ratio Test:

converges

= e^{3} converges

= converges

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What is the sum from n=2 to infinity of e^{3-6n} ?
The sum from n=2 to infinity of e^{3-6n} is converges