What is the sum from n=4 to infinity of (6^n)/(n!) ?

The sum from n=4 to infinity of (6^n)/(n!) is converges

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sum from n=4 to infinity of (6^n)/(n!)

n = 4 \sum \infty \frac{6 ^{n}}{n !}

converges

Solution steps

n = 4 \sum \infty \frac{6 ^{n}}{n !}

Apply Series Ratio Test:

converges

= converges

n = 2 \sum \infty \frac{1}{n} - \frac{1}{n - 1}

n = 1 \sum \infty \frac{1}{4 n + 1}

n = 1 \sum \infty (cos (8))^{n}

n = 1 \sum \infty \frac{1}{n ^{9} + 9}

n = 1 \sum \infty \frac{5}{n ^{2} + 4}

What is the sum from n=4 to infinity of (6^n)/(n!) ?
The sum from n=4 to infinity of (6^n)/(n!) is converges