What is the sum from n=0 to infinity of (6^n)/(n!) ?

The sum from n=0 to infinity of (6^n)/(n!) is converges

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sum from n=0 to infinity of (6^n)/(n!)

n = 0 \sum \infty \frac{6 ^{n}}{n !}

converges

Solution steps

n = 0 \sum \infty \frac{6 ^{n}}{n !}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty n^{2} e^{- 2 n}

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

n = 0 \sum \infty (1 - n)^{- 2}

n = 0 \sum \infty \frac{3}{2 n}

n = 0 \sum \infty \frac{1}{6} (\frac{5}{6})^{n}

What is the sum from n=0 to infinity of (6^n)/(n!) ?
The sum from n=0 to infinity of (6^n)/(n!) is converges