# Find the expression for the sum of this power series: $\sum_{n=0}^{\infty} \frac{x^{n}}{(n+3)!}$

I'm a bit stuck on this problem. I supposed to find an expression for the sum of this power series

$$\sum_{n=0}^{\infty} \frac{x^{n}}{(n+3)!}$$

I know I should use $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$

But then I get confused, and I don't how I shall proceed to find the answer.

• Hint: You know that $e^x = \sum\limits_{n=0}^\infty \dfrac{x^n}{n!}$. What is $\sum\limits_{n=\color{red}{3}}^\infty \dfrac{x^n}{n!}$ where we changed the lower limit to $n=3$ instead? How does this compare to $e^x$? How does this compare to your series? Commented May 15, 2020 at 12:02
• Hmm, I guess I could write $\sum_{n=3}^{\infty} \frac{x^{n+3}}{(n+3)!}=e^x$ and then mulitply by $1/x^3$ on both sides getting $\frac{e^x}{x^3}=\sum_{n=3}^{\infty} \frac{x^n}{(n+3)!}$. I'm I thinking right? Commented May 15, 2020 at 12:09
• Almost... multiplying by $\frac{1}{x^3}$ is certainly going to be useful here. But what you wrote isn't equal to $e^x$. Maybe for intuition try writing it out some terms. $e^x = 1+\frac{x}{1}+\frac{x^2}{2!}+\color{blue}{\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\dots}$. On the other hand you have $\sum\limits_{n=3}^\infty \frac{x^n}{n!} = \color{blue}{\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\dots}$ Commented May 15, 2020 at 12:12
• @JMoravitz I'm sorry, but I have a hard time understanding. I see now that what i wrote was wrong, I would have to write $\sum_{n=3}^{\infty} \frac{x^{n-3}}{(n-3)!}$ for it to be equal to $e^x$. But that confuses me even more :( Commented May 15, 2020 at 12:38
• Okay, I think I understand what you mean now (or I hope so). So $e^x=1+x+\frac{1}{x^2}+\sum_{n=3}^{\infty} \frac{x^n}{n!}$. Then must $e^x-1-x-\frac{x^2}{2!}=\sum_{n=3}^{\infty} \frac{x^n}{n!}$ But then I find it hard to proceed. I'm sorry for asking so much. I'm just curious when it comes to how you would solve it. Commented May 15, 2020 at 13:02

$$\sum_{n=0}^{\infty} \frac{x^{n}}{(n+3)!}=\sum_{n=3}^{\infty} \frac{x^{n-3}}{n!}=x^{-3}\sum_{n=3}^{\infty} \frac{x^{n}}{n!}=x^{-3}(e^x-\frac{x^0}{0!}-\frac{x^1}{1!}-\frac{x^2}{2!})=\frac{e^x}{x^3}-\frac{1}{x^3}-\frac{1}{x^2}-\frac{1}{2x}$$ I believe that I've done everything correctly.
• I get a bit lost when it comes to your fourth expression when you get rid of the sigma. Is not $\sum_{n=3}^{\infty} \frac{x^n}{n!}= \frac{x^3}{3!}+...$ Where does $e^x$ come from? Commented May 15, 2020 at 12:45
• $e^x=\sum_{n=0}^\infty{x^n/n!}$ therefore $\sum_{n=3}^\infty{x^n/n!}$ is equal to $e^x$ minus the first three terms which are $x^0/0!$, $x^1/1!$, $x^2/2!$ Commented May 15, 2020 at 12:52