What is the sum from n=0 to infinity of 1/(e^{4n)} ?

The sum from n=0 to infinity of 1/(e^{4n)} is converges

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sum from n=0 to infinity of 1/(e^{4n)}

n = 0 \sum \infty \frac{1}{e ^{4 n}}

converges

Solution steps

n = 0 \sum \infty \frac{1}{e ^{4 n}}

Apply Series Ratio Test:

converges

= converges

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

k = 0 \sum \infty (- \frac{1}{5})^{k}

n = 1 \sum \infty \frac{8 - 5}{n ^{2}}

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

n = 1 \sum \infty \frac{5}{17 + 8 ^{n}}

What is the sum from n=0 to infinity of 1/(e^{4n)} ?
The sum from n=0 to infinity of 1/(e^{4n)} is converges