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Popular >

sum from n=2 to infinity of (n+4)/(n!)

Solution

n = 2 \sum \infty \frac{n + 4}{n !}

Solution

converges

Solution steps

n = 2 \sum \infty \frac{n + 4}{n !}

Expand

\frac{n + 4}{n !} : \frac{n}{n !} + \frac{4}{n !}

= n = 2 \sum \infty \frac{n}{n !} + \frac{4}{n !}

Apply the Sum Rule:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

= n = 2 \sum \infty \frac{n}{n !} + n = 2 \sum \infty \frac{4}{n !}

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

converges

n = 2 \sum \infty \frac{4}{n !} =

converges

= converges + converges

= converges

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Frequently Asked Questions (FAQ)

What is the sum from n=2 to infinity of (n+4)/(n!) ?
The sum from n=2 to infinity of (n+4)/(n!) is converges