What is the sum from n=1 to infinity of (2n!)/(n^5) ?

The sum from n=1 to infinity of (2n!)/(n^5) is diverges

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sum from n=1 to infinity of (2n!)/(n^5)

n = 1 \sum \infty \frac{2 n !}{n ^{5}}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n !}{n ^{5}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{n !}{n ^{5}}

Apply Series Ratio Test:

diverges

= 2 diverges

= diverges

\frac{\partial}{\partial x} (- 6 e^{- 3 x^{2} - y^{2}} x)

(1 - 2 x) (3 x - 4), at x = 2

f (x) = 2 x^{2} - x + 9

f (x) = 2 - \frac{2}{3} x

\frac{d}{d x} (ln (ln (sec (2 x))))

What is the sum from n=1 to infinity of (2n!)/(n^5) ?
The sum from n=1 to infinity of (2n!)/(n^5) is diverges