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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (x^{2} e^{y})

\int s 4^{s} d s

\int_{0}^{\frac{π}{2}} (cos (x))^{2} d x

n = 1 \sum \infty 4^{\frac{1}{2 ^{n}}}

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