What is the sum from n=0 to infinity of n!2^nn^{-n} ?

The sum from n=0 to infinity of n!2^nn^{-n} is converges

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sum from n=0 to infinity of n!2^nn^{-n}

n = 0 \sum \infty n! 2^{n} n^{- n}

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 8

y = \frac{x ^{x}}{sin ( x )}

x \to 2 lim (\frac{x ^{3} + 8}{x - 2})

\frac{\partial}{\partial y} (e^{x - 1} cos (y))

\frac{\partial}{\partial x} (72 y^{7} x^{7} - 56 y^{6} x^{6})

What is the sum from n=0 to infinity of n!2^nn^{-n} ?
The sum from n=0 to infinity of n!2^nn^{-n} is converges