Find the sum of the series $\sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}$ Find the sum of the series $\sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}$ using the Taylor series of $e^{x+2 }$. 
Answer:
$$
e^{x+2}=1+(x+2)+\frac{(x+2)^2}{2!}+\frac{(x+2)^3}{3!}+\ldots
$$
Integrating, we get
$$
\int e^{x+2} dx=x+\frac{(x+2)^2}{2!}+\frac{(x+2)^3}{3!}+\frac{(x+2)^4}{4!}+\ldots
$$
Again integrating, we get
$$
\int e^{x+2} dx=\frac{x^2}{2}+\frac{(x+2)^3}{3!}+\frac{(x+2)^4}{4!}+\frac{(x+2)^5}{5!}+\ldots\\
\Rightarrow \int e^{x+2} dx=\frac{x^2}{2}+(x+2)^3 \sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}\\
\Rightarrow \sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}=\frac{\int e^{x+2} dx-\frac{x^2}{2}}{(x+2)^3}
$$
Is this the sum of the series ?

Your approach is sound, but you should be clear about one thing: when you write $\int e^{x+2}\,\mathrm dx$, what you mean is $\int_0^xe^{t+2}\,\mathrm dt$.
