New Year Countdown - 3

First of all,let us notice the terms in the numerator and the difference between the consecutive terms $\\ 3,5,9,15,23\\ \ \ 2\ \ 4\ \ 6\ \ \ 8$ .As you can see,the differences make an $A.P$ and to get the $n^{th}$ term of the sequence,you have to do this, $3(\text{first term})+\text{sum of }(n-1)\text{terms of the A.P}(2,4,6,8....).\\ =3+\dfrac{n-1}{2}(2+2+(n-2)*2)\\ =3+n*(n-1).$ So, $n^{th}$ term of the given series $=\dfrac{3+n(n-1)}{n!}$ .So,we have to find, ${\displaystyle\sum_{n={1}}^{\infty}(\dfrac{3+n(n-1)}{n!})}\\ =\displaystyle\sum_{n={1}}^{\infty}(\dfrac{3}{n!}+\dfrac{1}{(n-2)!})\\ =\displaystyle\sum_{n={1}}^{\infty}(\dfrac{3}{n!})+(\displaystyle\sum_{n={1}}^{\infty}\dfrac{1}{(n-2)!}).$ The first part of the expression is $3*(e-1)$ where $e$ is the Euler's constant.Now,for calculating the second part of the expression,first notice this, $\sum_{n=2}^{\infty}(\dfrac{1}{(n-2)!})=e$ .So we just have to calculate the value of $\dfrac{1}{(-1)!}$ and we are done.Now, $(-1)!=\infty$ .Thus, $\dfrac{1}{(-1)!}=0.$ Thus,the value of $\displaystyle\sum_{n={1}}^{\infty}\dfrac{1}{(n-2)!}=0+e=e$ .So,our answer $=3*(e-1)+e\\ =7.87.$

3(e-1) + 1 = ans.