What is the sum from n=0 to infinity of (n^2+1)/(n!) ?

The sum from n=0 to infinity of (n^2+1)/(n!) is converges

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sum from n=0 to infinity of (n^2+1)/(n!)

n = 0 \sum \infty \frac{n ^{2} + 1}{n !}

converges

Solution steps

n = 0 \sum \infty \frac{n ^{2} + 1}{n !}

Expand

\frac{n ^{2} + 1}{n !} : \frac{n ^{2}}{n !} + \frac{1}{n !}

= n = 0 \sum \infty \frac{n ^{2}}{n !} + \frac{1}{n !}

Apply the Sum Rule:

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

= n = 0 \sum \infty \frac{n ^{2}}{n !} + n = 0 \sum \infty \frac{1}{n !}

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

converges

n = 0 \sum \infty \frac{1}{n !} =

converges

= converges + converges

= converges

What is the sum from n=0 to infinity of (n^2+1)/(n!) ?
The sum from n=0 to infinity of (n^2+1)/(n!) is converges