# Series $\frac{x^{3n}}{(3n)!}$ find sum using differentiation

Find sum of the series $$\sum_{n=1}^{\infty}\frac{x^{3n}}{\left(3n\right)!}$$ using differentiation. So far I found that $$S(x)+1=S'''(x)$$ but it does not help. Then I tried different interesting ideas like $$S(x)+S'(x)+S''(x)=e^x-1\,.$$Maybe if I get the third equation it will allow me to construct a kind of differential equation. Then, by solving it, obtain $S(x)$.

• The magic word is multisection. – marty cohen May 24 '16 at 18:24

## 3 Answers

Define $$y(x)=\sum_{k\geq0}\frac{x^{3k}}{\left(3k\right)!}.$$ We observe that $$\sum_{k\geq0}\frac{x^{3k}}{\left(3k\right)!}+\sum_{k\geq0}\frac{x^{3k+1}}{\left(3k+1\right)!}+\sum_{k\geq0}\frac{x^{3k+2}}{\left(3k+2\right)!}=e^{x}$$ so we have the second order ODE $$y''\left(x\right)+y'\left(x\right)+y\left(x\right)=e^{x}.\tag{1}$$ Let start to solve the characteristic polynomial $$\lambda^{2}+\lambda+1=0\Leftrightarrow\lambda_{1,2}=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$$ so a solution of the homogeneous equation is $$y_{o}\left(x\right)=c_{1}e^{-x/2}\sin\left(\frac{\sqrt{3}x}{2}\right)+c_{2}e^{-x/2}\cos\left(\frac{\sqrt{3}x}{2}\right),\, c_{1},c_{2}\in\mathbb{R}.$$ Now we have to found a particolar solution. Since the known term is $e^{x}$ we know that a particular solution will be of the form $$y_{p}\left(x\right)=Ae^{x}\tag{2}$$ where $A\in\mathbb{R}$. So substituting $(2)$ in $(1)$ and equaling the coefficients, we get $$A=\frac{1}{3}$$ so, since $y\left(0\right)=1,\,y'\left(0\right)=0$ we have $$y\left(x\right)=y_{o}\left(x\right)+y_{p}\left(x\right)=\frac{2}{3}e^{-x/2}\cos\left(\frac{\sqrt{3}x}{2}\right)+\frac{e^{x}}{3}$$ and finally $$\sum_{k\geq1}\frac{x^{3k}}{\left(3k\right)!}=\frac{2}{3}e^{-x/2}\cos\left(\frac{\sqrt{3}x}{2}\right)+\frac{e^{x}}{3}-1.$$

• Why $y\left(0\right)=y'\left(0\right)=0$ => $y(x) = 2/3 e^{-x/2} cos(\sqrt(3)x/2)+e^x/3$ ? $y\left(x\right)=y_{o}\left(x\right)+y_{p}\left(x\right)=\frac{2}{3}e^{-0/2}\cos\left(\frac{\sqrt{3}\times0}{2}\right)+\frac{e^{0}}{3}=2/3+1/3 = 1 \neq 0$? – Zack Ni Jul 7 '16 at 12:08
• @ZackNi There was a mistake, thank you. I calculated $y(0)-1$ instead of $y(0).$ – Marco Cantarini Jul 7 '16 at 15:22

Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. Then $\frac{\omega^n+\omega^{2n}+1}{3}$ equals $1$ iff $3\mid n$, hence:

$$\sum_{n\geq 1}\frac{x^{3n}}{(3n)!} = \color{red}{-1+\frac{e^{x}+e^{\omega x}+e^{\omega^2 x}}{3}}. \tag{1}$$

• Not that it's a mistake or anything, but I don't see OP asking for the value of the series at $1$. Also, OP specifies they want the summation to start from $n=1$. – Wojowu May 22 '16 at 15:47
• He is not giving the value at 1. He uses that expression to produce the desired series from the exponential series. It is called multisection. – marty cohen May 24 '16 at 18:27


\begin{align} \color{#f00}{\sum_{n = 1}^{\infty}{x^{3n} \over \pars{3n}!}} & = -1 + \sum_{n = 0}^{\infty}{x^{n} \over n!}\sum_{k = 0}^{\infty}\delta_{3k,n} = -1 + \sum_{n = 0}^{\infty}{x^{n} \over n!}\sum_{k = 0}^{\infty} \oint_{\verts{z} = a}{1 \over z^{n - 3k + 1}}\,{\dd z \over 2\pi\ic} \quad\mbox{where}\quad 0 < a < 1 \end{align}

Then, \begin{align} \color{#f00}{\sum_{n = 1}^{\infty}{x^{3n} \over \pars{3n}!}} & = -1 + \oint_{\verts{z} = a}\sum_{k = 0}^{\infty}z^{3k - 1} \sum_{n = 0}^{\infty}{\pars{x/z}^{n} \over n!}\,{\dd z \over 2\pi\ic} = -1 + \oint_{\verts{z} = a} {\expo{x/z} \over z\pars{1 - z^{3}}}\,{\dd z \over 2\pi\ic} \\[3mm] & = -1 + \oint_{\verts{z} = 1/a}{z^{2}\expo{xz} \over z^{3} - 1} \,{\dd z \over 2\pi\ic} = -1 + {1 \over 3}\sum_{r}\expo{xr}\quad\mbox{where}\quad r \in \braces{1,\expo{-2\pi\ic/3},\expo{2\pi\ic/3}} \end{align}

$r$'s are roots of $z^{3} - 1 = 0$ whith $\verts{r} = 1 < 1/a > 1$: \begin{align} \color{#f00}{\sum_{n = 1}^{\infty}{x^{3n} \over \pars{3n}!}} & = -1 + {1 \over 3}\,\expo{x} + {2 \over 3}\, \Re\pars{\exp\pars{x\bracks{-\,\half + {\root{3} \over 2}\,\ic}}} \\[3mm] & = \color{#f00}{% -1 + {1 \over 3}\,\expo{x} + {2 \over 3}\, \expo{-x/2}\cos\pars{{\root{3} \over 2}\,x}} \end{align}