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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Ratio Test:

converges

= 2 converges

= converges

\int \frac{6 x ^{3} + 2 x ^{2} - 47 x - 9}{x ^{2} - 9} d x

f (x) = 7 x - 2 x^{2}, (- 1, - 9)

\int (2 x + 7)^{2} d x

\int \frac{1}{1 - sin ( 2 x )} d x

(1 - 2 t)^{- 4} (t^{2} - t)^{2}

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