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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{\partial}{\partial x} (2 x ln (x y))

\int_{0}^{50} 0.624 x d x

3 x, 4 - x^{2}

3 e^{2 x} (\frac{d y}{d x}) = - 4 (\frac{x}{y ^{2}})

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