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

n = 1 \sum \infty (- 1)^{n + 1} \cdot n

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

n = 1 \sum \infty cos (\frac{7}{n})

n = 0 \sum \infty \frac{1}{4 + 3 n}

n = 2 \sum \infty \frac{3}{n ln ^{9} ( 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