What is the sum from n=0 to infinity of (9n!)/(n^n) ?

The sum from n=0 to infinity of (9n!)/(n^n) is converges

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sum from n=0 to infinity of (9n!)/(n^n)

n = 0 \sum \infty \frac{9 n !}{n ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{9 n !}{n ^{n}}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{n !}{n ^{n}}

Apply Series Ratio Test:

converges

= 9 converges

= converges

\frac{d}{d y} (\frac{1}{y ^{3} + 4 y ^{2}})

n = 0 \sum \infty 2^{- n}

y^{^{''}} + 4 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int (\frac{( ln ( x ) ) ^{3}}{x}) d x

\int_{0}^{4} sin (x) d x

What is the sum from n=0 to infinity of (9n!)/(n^n) ?
The sum from n=0 to infinity of (9n!)/(n^n) is converges