What is the sum from n=5 to infinity of (1^n)/(n!) ?

The sum from n=5 to infinity of (1^n)/(n!) is converges

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sum from n=5 to infinity of (1^n)/(n!)

n = 5 \sum \infty \frac{1 ^{n}}{n !}

converges

Solution steps

n = 5 \sum \infty \frac{1 ^{n}}{n !}

Apply Series Ratio Test:

converges

= converges

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

n = 1 \sum \infty (\frac{6}{9})^{n - 1}

n = 0 \sum \infty 3 (1 - n)^{- 2}

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

n = 0 \sum \infty (\frac{4}{11})^{n}

What is the sum from n=5 to infinity of (1^n)/(n!) ?
The sum from n=5 to infinity of (1^n)/(n!) is converges