What is the sum from n=0 to infinity of (5n!)/(2^n) ?

The sum from n=0 to infinity of (5n!)/(2^n) is diverges

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sum from n=0 to infinity of (5n!)/(2^n)

n = 0 \sum \infty \frac{5 n !}{2 ^{n}}

diverges

Solution steps

n = 0 \sum \infty \frac{5 n !}{2 ^{n}}

Apply the constant multiplication rule:

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

= 5 \cdot n = 0 \sum \infty \frac{n !}{2 ^{n}}

Apply Series Ratio Test:

diverges

= 5 diverges

= diverges

n = 0 \sum \infty \frac{5}{6 ^{n}}

n = 2 \sum \infty (\frac{3}{8})^{4 n}

n = 0 \sum \infty \frac{4}{( 1 - 4 n ) ^{2}}

n = 1 \sum \infty n^{5} e^{n^{3}}

n = 2 \sum \infty (\frac{3}{5})^{2 n}

What is the sum from n=0 to infinity of (5n!)/(2^n) ?
The sum from n=0 to infinity of (5n!)/(2^n) is diverges