Find the value of Σk=0^∞(k^2+k+1)/((k+2)!).

Write the numerator in terms of $m=k+2$: $k^{2}+k+1=(m-2)^{2}+(m-2)+1=m^{2}-3m+3$. The sum becomes $\sum_{m=2}^{\infty}\frac{m^{2}-3m+3}{m!}$. Now $\frac{m^{2}}{m!}=\frac{1}{(m-2)!}+\frac{1}{(m-1)!}$ and $\frac{m}{m!}=\frac{1}{(m-1)!}$, so $\frac{m^{2}-3m+3}{m!}=\frac{1}{(m-2)!}-\frac{2}{(m-1)!}+\frac{3}{m!}$. Summing from $m=2$ gives $e-2(e-1)+3(e-2)=2e-4$. Numerically $2e-4=1.43656$, and a direct partial sum of the original series agrees. The answer rounds to $1.44$. The factorial shift is the main idea: shifting the index to $m=k+2$ lets us express everything through shifted exponential series. Series of the form $\sum \frac{P(m)}{m!}$ with a polynomial numerator always sum to a combination $re+s$ with rational $r,s$, so the final value here is $2e-4$ without any numerical approximation.