<s>[INST] 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 ?
 [/INST] There is no need for integrals.
Multiply the sum by $(x+2)^3$ to let the exponents match the denominators and add the missing initial terms to obtain an exponential:
$$1+(x+2)+\frac{(x+2)^2}2+(x+2)^3S=\sum_{n=-3}^\infty\frac{(x+2)^{n+3}}{(n+3)!}=e^{x+2}$$
from which you draw $S$ (which is not what you found).
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