What is the sum from n=5 to infinity of e^{5-6n} ?

The sum from n=5 to infinity of e^{5-6n} is converges

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sum from n=5 to infinity of e^{5-6n}

n = 5 \sum \infty e^{5 - 6 n}

converges

Solution steps

n = 5 \sum \infty e^{5 - 6 n}

Simplify

e^{5 - 6 n} : e^{5} e^{- 6 n}

= n = 5 \sum \infty e^{5} e^{- 6 n}

Apply the constant multiplication rule:

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

= e^{5} \cdot n = 5 \sum \infty e^{- 6 n}

Apply Series Integral Test:

converges

= e^{5} converges

= converges

n = 0 \sum \infty \frac{1}{n ^{10}}

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

n = 1 \sum \infty π

n = 1 \sum \infty ln (n - 1)

n = 1 \sum \infty \frac{( - 5 ) ^{n}}{4}

What is the sum from n=5 to infinity of e^{5-6n} ?
The sum from n=5 to infinity of e^{5-6n} is converges