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

The sum from n=1 to infinity of (-1)^n(2n!)/(2^nn!n) is converges

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty (- 1)^{n} \frac{n !}{2 ^{n} n ! n}

Apply Alternating Series Test:

converges

= 2 converges

= converges

What is the sum from n=1 to infinity of (-1)^n(2n!)/(2^nn!n) ?
The sum from n=1 to infinity of (-1)^n(2n!)/(2^nn!n) is converges