What is the sum from n=1 to infinity of (6n^2)/(n!) ?

The sum from n=1 to infinity of (6n^2)/(n!) is converges

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sum from n=1 to infinity of (6n^2)/(n!)

n = 1 \sum \infty \frac{6 n ^{2}}{n !}

converges

Solution steps

n = 1 \sum \infty \frac{6 n ^{2}}{n !}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{n ^{2}}{n !}

Apply Series Ratio Test:

converges

= 6 converges

= converges

n = 1 \sum \infty - 7 (\frac{3}{8})^{n}

n = 1 \sum \infty \frac{1}{4 + 3 n}

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

n = 1 \sum \infty (\frac{1}{2})^{n - 2}

k = 1 \sum \infty 6 (\frac{1}{5})^{k - 1}

What is the sum from n=1 to infinity of (6n^2)/(n!) ?
The sum from n=1 to infinity of (6n^2)/(n!) is converges