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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

x = 1 \sum \infty x

k = 2 \sum \infty \frac{1}{k + 1} - \frac{1}{k}

n = 0 \sum \infty \frac{5}{2 ^{n - 1}}

n = 0 \sum \infty \frac{n}{1 - 4 n}

n = 16 \sum \infty \frac{1}{ln ( n )}

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