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 = 0 \sum \infty ln (3 - n)

n = 0 \sum \infty \frac{1}{3 ^{n} - 1}

n = 0 \sum \infty 0.52 (0.01)^{n}

n = 1 \sum \infty 2 + \frac{cos ( n )}{n}

n = 0 \sum \infty (\frac{13}{3})^{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