Evaluate the following sum: $$\dfrac{1}{1!+2!+3!}+\dfrac{1}{2!+3!+4!}+\dots + \dfrac{1}{2016!+2017!+2018!}$$
I was trying to rewrite the general term as: $$\frac{1}{n!+(n+1)!+(n+2)!}=\frac{1}{n!(n+2)^2}$$
However, this did not give any essential improvements.
On
$$\sum_{k = 0}^{+\infty} \frac{1}{n! + (n+1)! + (n+2)!} = -\text{Ei}(1)+\gamma +e-1$$
Where $Ei$ is the Exponential Integral special function, and $\gamma$ is the Euler-Mascheroni constant.
More generally we have:
$$\sum_{k = 0}^{N} \frac{1}{n! + (n+1)! + (n+2)!} = $$
$$ = \frac{(-\text{Ei}(1)+\gamma +e-1) ((N+1)!+(N+2)!+(N+3)!)-\, _3F_3(1,N+3,N+3;N+2,N+4,N+4;1)}{(N+1)!+(N+2)!+(N+3)!}$$
Where you are facing the Hypergeometric special function.
Using the exponential integral function $\text{Ei}$, one can show that $$\sum_{n=1}^\infty\frac{x^{n+2}}{n!(n+2)^2}=e^x-\frac14x^2-1+\ln x-\text{Ei}(x)+\gamma$$ hence $$\sum_{n=1}^\infty\frac{1}{n!(n+2)^2}=e-\frac54-\text{Ei}(1)+\gamma$$ Since there is no expression of $\text{Ei}(1)$ except as the sum of a series quite related to the infinite series corresponding to the series in the question, this mainly shows the question has no satisfying answer.