# calculate $\sum_{n=0}^\infty \frac{3^n}{n!(n+3)}$ using power series

let $$f(x)=\frac{e^x-1-x-\frac{x^2}{2}}{x}$$, because $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$, $$f$$ can be expressed as $$f(x) = \frac{\sum_{n=0}^\infty \frac{x^n}{n!}-1-x-\frac{x^2}{2}}{x}=\frac{\sum_{n=3}^\infty \frac{x^n}{n!}}{x}=\sum_{n=0}^\infty \frac{x^{n+2}}{(n+3)!}$$ the power series converge in $$(-\infty, \infty)$$ because $$\lim_{n\to\infty} \sqrt[n]{\frac{1}{(n+3)!}}=0$$ and let $$f_n(x) = \frac{x^{n+2}}{(n+3)!} \Longrightarrow f'_n(x) = \frac{x^{n+1}}{(n+1)!(n+3)}$$, $$\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)!(n+3)}= \sum_{n=0}^\infty f'_n(x)$$ also converge in $$(-\infty, \infty)$$ (for the same reason), hence $$f'(x) = \sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)!(n+3)}$$ by repeating this process once more I get $$f''(x) = \sum_{n=0}^\infty \frac{x^n}{n!(n+3)}$$ and if $$x=3$$ I get $$\sum_{n=0}^\infty \frac{3^n}{n!(n+3)} = f''(3)$$ which is what was looking for. my problem is that $$f$$ isn't defined for $$x=0$$ yet the series does converge for it as $$\sum_{n=0}^\infty \frac{0^n}{n!(n+3)}=0$$, so was the function $$f$$ I used wrong? or could it be that I can't differentiate $$f$$ the way I did?

Hint

$$f$$ is not formally defined at $$0$$. However you can extend it by continuity at $$0$$.

In particular

$$\lim\limits_{x \to 0} \frac{e^x - 1}{x} = (e^x)^\prime(0) = 1$$

Hence you can extend $$f$$ by continuity at $$0$$ with $$f(0)=0$$.

There is no contradiction and what you did seems OK regarding the computations.

You are correct that the version of $$f$$ given by the recipe $$\frac{\mathrm{e}^x-1-x-x^2/2}{x}$$ is undefined at $$x = 0$$. However, the limit of this recipe as $$x \rightarrow 0$$ is $$0$$, so there is a continuous function, $$\hat{f}$$, with domain $$(-\infty, \infty)$$, which agrees with $$f$$ on $$(-\infty, \infty) \smallsetminus \{0\}$$ and agrees with $$f$$'s limit as $$x \rightarrow 0$$. You have already written a recipe for $$\hat{f}$$, when you wrote $$\sum_{n=0}^\infty x^{n+2}/(n+3)!$$.

Since you manipulated the series, $$\hat{f}$$, with infinite radius of convergence, you need not worry about the defects of the original recipe. Furthermore, $$f$$ and $$\hat{f}$$ agree at $$x = 3$$, the only point at which you intend to evaluate. You would be properly concerned if you were trying to evaluate at $$x = 0$$, but you are not.

Try using the following instead so you avoid that mess: $$x^2e^x=\sum_{n=0}^{\infty} \frac{x^{n+2}}{n!}$$

$$\int x^2e^x \; dx=x^2e^x-2xe^x+2e^x+C=\sum_{n=0}^{\infty} \frac{x^{n+3}}{n!(n+3)}$$ At $$x=3$$: $$9e^3-6e^3+2e^3-2=5e^3-2=\sum_{n=0}^{\infty} \frac{3^{n+3}}{n!(n+3)}$$

Notice that the series on the right is what you're looking for but multiplied by $$3^3$$, so divide both sides by $$27$$.

I understand that this approach is slightly different than yours, but I believe this approach is faster and easier to understand so I thought you might appreciate it. I'm sure you can refer to other answers posted here regarding your confusion with $$f(0)$$.

$$\boxed{\frac{5e^3-2}{27}}$$

• No aspect of this Answer responds to the Question's "was the function $f$ I used wrong? or could it be that I can't differentiate $f$ the way I did?" The method proposed here is subject to exactly the same questions. May 30, 2020 at 16:07
• The function is defined at $x=0$ in my answer. I was showing an alternate and easier way to calculate the sum.
– Ty.
May 30, 2020 at 16:09
• while this way is, by far, more elegant, I haven't "learnt" integrals with series yet, thus I had to use derivative. May 30, 2020 at 16:37