Find the sum of the series $\sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}$

Question

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 ?

Answer 1

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 $\int_0^xe^{t+2}\,\mathrm dt$.

Answer 2

As $\displaystyle e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$

Set $n+3=m$

$$\sum_{n=0}^\infty\dfrac{(x+2)^n}{(n+3)!}=\sum_{m=3}^\infty\dfrac{(x+2)^{m-3}}{m!}=\dfrac1{(x+2)^3}\sum_{m=3}^\infty\dfrac{(x+2)^m}{m!}$$

Now, $$\sum_{m=3}^\infty\dfrac{(x+2)^m}{m!}=e^{x+2}-\sum_{m=0}^2\dfrac{(x+2)^m}{m!}$$

Answer 3

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).