What is the sum from n=1 to infinity of (3n!)/(n^n) ?

The sum from n=1 to infinity of (3n!)/(n^n) is converges

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sum from n=1 to infinity of (3n!)/(n^n)

n = 1 \sum \infty \frac{3 n !}{n ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{3 n !}{n ^{n}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{n !}{n ^{n}}

Apply Series Ratio Test:

converges

= 3 converges

= converges

\int_{0}^{\frac{π}{2}} sin (2 x) d x

3 t^{4} - 2 t^{2} + 1

cos (2 t)

y = 16 - x^{2}

\frac{\partial}{\partial x} ((a - x)^{2} + b (y - x^{2})^{2})

What is the sum from n=1 to infinity of (3n!)/(n^n) ?
The sum from n=1 to infinity of (3n!)/(n^n) is converges