What is the sum from n=0 to infinity of (2n^2)/(n!) ?

The sum from n=0 to infinity of (2n^2)/(n!) is converges

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sum from n=0 to infinity of (2n^2)/(n!)

n = 0 \sum \infty \frac{2 n ^{2}}{n !}

converges

Solution steps

n = 0 \sum \infty \frac{2 n ^{2}}{n !}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty \frac{n ^{2}}{n !}

Apply Series Ratio Test:

converges

= 2 converges

= converges

What is the sum from n=0 to infinity of (2n^2)/(n!) ?
The sum from n=0 to infinity of (2n^2)/(n!) is converges