What is the sum from n=1 to infinity of (6n!)/(n^n) ?

The sum from n=1 to infinity of (6n!)/(n^n) is converges

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sum from n=1 to infinity of (6n!)/(n^n)

n = 1 \sum \infty \frac{6 n !}{n ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{6 n !}{n ^{n}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{n !}{n ^{n}}

Apply Series Ratio Test:

converges

= 6 converges

= converges

\frac{d}{d x} (\frac{x ^{2} - 5}{1 - 3 x ^{2}})

6 (sin (x^{2}) + 2 x^{2} cos (x^{2}))

\frac{d}{d x} (\frac{x ^{2} tan ( x ^{3} - x ^{2} )}{sec ( x )})

y^{^{''}} + 3 y^{^{'}} = 10 sin (t)

y = 5 x - 4 x, (1, 1)

What is the sum from n=1 to infinity of (6n!)/(n^n) ?
The sum from n=1 to infinity of (6n!)/(n^n) is converges