What is the sum from n=1 to infinity of 1/(n!4^n) ?

The sum from n=1 to infinity of 1/(n!4^n) is converges

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sum from n=1 to infinity of 1/(n!4^n)

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

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ! \cdot 4 ^{n}}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{1}{3 - n}

n = 1 \sum \infty \frac{3}{2 n}

k = 1 \sum \infty k e^{- 7 k}

n = 0 \sum \infty 2 (\frac{3}{2})^{n}

k = 0 \sum \infty (\frac{2}{3})^{k}

What is the sum from n=1 to infinity of 1/(n!4^n) ?
The sum from n=1 to infinity of 1/(n!4^n) is converges