What is the sum from n=0 to infinity of (7n!)/(2^n) ?

The sum from n=0 to infinity of (7n!)/(2^n) is diverges

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sum from n=0 to infinity of (7n!)/(2^n)

n = 0 \sum \infty \frac{7 n !}{2 ^{n}}

diverges

Solution steps

n = 0 \sum \infty \frac{7 n !}{2 ^{n}}

Apply the constant multiplication rule:

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

= 7 \cdot n = 0 \sum \infty \frac{n !}{2 ^{n}}

Apply Series Ratio Test:

diverges

= 7 diverges

= diverges

f (x) = 1 - x - x^{2}, at x = 0

x (29 - 2 x) (16 - 2 x)

lo g_{7} (x)

ln (x + 2), - 1

2 x y^{^{'}} - y = 3 x^{2}

What is the sum from n=0 to infinity of (7n!)/(2^n) ?
The sum from n=0 to infinity of (7n!)/(2^n) is diverges