What is the sum from n=1 to infinity of (2n!)/(n^n) ?

The sum from n=1 to infinity of (2n!)/(n^n) is converges

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sum from n=1 to infinity of (2n!)/(n^n)

n = 1 \sum \infty \frac{2 n !}{n ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{2 n !}{n ^{n}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2 \cdot n = 1 \sum \infty \frac{n !}{n ^{n}}

Apply Series Ratio Test:

converges

= 2 converges

= converges

r (2 r + 1)

\frac{d y}{d t} = 5 (y - t^{2})

\int \frac{x ^{3}}{( 2 + x ^{4} ) ^{9}} d x

y^{^{'}} = 2 y - 3 e^{- t}

x \to 0 lim (sin (\frac{2}{x}))

What is the sum from n=1 to infinity of (2n!)/(n^n) ?
The sum from n=1 to infinity of (2n!)/(n^n) is converges