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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

diverges

= \frac{1}{2} diverges

= diverges

n = 1 \sum \infty n^{- 0.6}

n = 0 \sum \infty \frac{4}{9 - n ^{2}}

k = 1 \sum \infty \frac{3}{k} - \frac{3}{k + 2}

n = 1 \sum \infty \frac{6 n}{6 n + 1}

n = 1 \sum \infty \frac{1}{n} - \frac{2}{3 n - 1}

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