What is the sum from n=1 to infinity of (9n^2)/(n!) ?

The sum from n=1 to infinity of (9n^2)/(n!) is converges

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sum from n=1 to infinity of (9n^2)/(n!)

n = 1 \sum \infty \frac{9 n ^{2}}{n !}

converges

Solution steps

n = 1 \sum \infty \frac{9 n ^{2}}{n !}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{n ^{2}}{n !}

Apply Series Ratio Test:

converges

= 9 converges

= converges

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

k = 1 \sum \infty (cos (6))^{k}

n = 1 \sum \infty \frac{n !}{8 6 ^{n}}

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

k = 1 \sum \infty k (\frac{7}{8})^{k}

What is the sum from n=1 to infinity of (9n^2)/(n!) ?
The sum from n=1 to infinity of (9n^2)/(n!) is converges