What is the sum from n=1 to infinity of (25n!)/(n^n) ?

The sum from n=1 to infinity of (25n!)/(n^n) is converges

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sum from n=1 to infinity of (25n!)/(n^n)

n = 1 \sum \infty \frac{25 n !}{n ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{25 n !}{n ^{n}}

Apply the constant multiplication rule:

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

= 25 \cdot n = 1 \sum \infty \frac{n !}{n ^{n}}

Apply Series Ratio Test:

converges

= 25 converges

= converges

\frac{d}{d x} (\frac{x}{f ( x )})

2 y y^{^{'}} - x y^{^{'}} = y - 2 x

\int (2 y sin (x) - x^{3} + ln (y)) d y

y^{^{''}} - 9 y^{^{'}} = 81 t

\frac{d}{d x} (\frac{arccos ( x )}{x})

What is the sum from n=1 to infinity of (25n!)/(n^n) ?
The sum from n=1 to infinity of (25n!)/(n^n) is converges