What is the sum from k=0 to infinity of (4^k)/(k!) ?

The sum from k=0 to infinity of (4^k)/(k!) is converges

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sum from k=0 to infinity of (4^k)/(k!)

k = 0 \sum \infty \frac{4 ^{k}}{k !}

converges

Solution steps

k = 0 \sum \infty \frac{4 ^{k}}{k !}

Apply Series Ratio Test:

converges

= converges

j = 1 \sum \infty 3 (\frac{1}{4})^{j}

n = 1 \sum \infty \frac{n}{n + 8}

n = 9 \sum \infty \frac{1}{9 n ^{4} + 8}

n = 0 \sum \infty (0. 2^{n}) 0.7

n = 0 \sum \infty (0. 2^{n}) 0.8

What is the sum from k=0 to infinity of (4^k)/(k!) ?
The sum from k=0 to infinity of (4^k)/(k!) is converges