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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty (\frac{11 π}{e})^{n}

n = 1 \sum \infty n^{- 0.9999}

n = 0 \sum \infty \frac{cos ( πn )}{n}

k = 0 \sum \infty \frac{4 ^{k}}{k !}

j = 1 \sum \infty 3 (\frac{1}{4})^{j}

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