What is the sum from n=2 to infinity of (n!)/(n^n) ?

The sum from n=2 to infinity of (n!)/(n^n) is converges

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sum from n=2 to infinity of (n!)/(n^n)

n = 2 \sum \infty \frac{n !}{n ^{n}}

converges

Solution steps

n = 2 \sum \infty \frac{n !}{n ^{n}}

Apply Series Ratio Test:

converges

= converges

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

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

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

n = 5 \sum \infty \frac{4 ^{n}}{5 ^{n}}

n = 1 \sum \infty \frac{n + 6}{n ^{2}}

What is the sum from n=2 to infinity of (n!)/(n^n) ?
The sum from n=2 to infinity of (n!)/(n^n) is converges