What is the sum from n=1 to infinity of (9n^7)/(n!) ?

The sum from n=1 to infinity of (9n^7)/(n!) is converges

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sum from n=1 to infinity of (9n^7)/(n!)

n = 1 \sum \infty \frac{9 n ^{7}}{n !}

converges

Solution steps

n = 1 \sum \infty \frac{9 n ^{7}}{n !}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{n ^{7}}{n !}

Apply Series Ratio Test:

converges

= 9 converges

= converges

\frac{d x}{d t} = - (x^{2} - x), x (0) = 2

\int_{0}^{3} ∣ 2 x - 3 ∣ d x

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 2, y^{^{'}} (0) = 2

5 x^{2} - 11 x - \frac{6}{x ^{2}}

\frac{d}{d x} (\frac{x ^{2} + 8 x + 3}{x})

What is the sum from n=1 to infinity of (9n^7)/(n!) ?
The sum from n=1 to infinity of (9n^7)/(n!) is converges