Find the sum of the infinite series [1+(5/1!)+(8/2!)+(11/3!)+.....]

Question

Excluding the 1st term of the series 1, if we start from the 2nd term-($\frac{5}{1!}$), I can locate that the numerators are in A.P with common difference 3, & 1st term 5.

Whereas the denominators are of the form n! where $n \in [1,\infty)$.So, the n th term is $[\frac{5+(n-1)3}{n!}]$.

But this is given as an infinite series & I am unable to solve this factorial-A.P combination.

Please Help

Thank You

Answer 1

$$1 + 5/1! + 8/2! + \ldots = -1 + \sum_{n =0}^{\infty} \frac{5 + (3n -3)}{n!} = -1 + 5 \sum_{n=0}^{\infty} \frac{1}{n!} = 5e - 1$$