Find the sum of the series using Taylor expansion

Question

Find the following sum

$$ \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}, x \in \Bbb R $$

Answer 1

(This is an expansion on my comment above.)

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

Using the ratio test, we have

$$\lim_{n\to\infty} \left|\frac{x^{3(n+1)}}{(3(n+1))!} \div \frac{x^{3n}}{(3n)!} \right| = \lim_{n\to\infty} \frac{|x|^3}{(3n+3)(3n+2)(3n+1)} = 0 < 1,$$ regardless of the value of $x$. Thus, the series converges everywhere. As usual, with power series, the resulting series is infinitely differentiable, and derivatives can be computed termwise. In particular, \begin{align*} f'(x) &= \sum_{n=0}^\infty 3n\frac{x^{3n - 1}}{(3n)!} = \sum_{n=1}^\infty 3n\frac{x^{3n - 1}}{(3n)!} = \sum_{n=1}^\infty \frac{x^{3n - 1}}{(3n - 1)!} = \sum_{n=0}^\infty \frac{x^{3n + 2}}{(3n + 2)!} \\ f''(x) &= \sum_{n=0}^\infty \frac{x^{3n + 1}}{(3n + 1)!} \\ f'''(x) &= \sum_{n=0}^\infty \frac{x^{3n}}{(3n)!} = f(x). \end{align*} In particular, $f'''(x) = f(x)$. We also have, \begin{align*} f(0) &= 1 + \frac{0}{3!} + \frac{0}{6!} + \ldots = 1 \\ f'(0) &= \frac{0}{2!} + \frac{0}{5!} + \frac{0}{8!} + \ldots = 0 \\ f''(0) &= \frac{0}{1!} + \frac{0}{4!} + \frac{0}{7!} + \ldots = 0. \end{align*} So, our function is the (necessarily unique) solution to the IVP: $$y''' - y = 0, \, y(0) = 1, y'(0) = 0, y''(0) = 0$$ The usual way to solve such problems is to use the characteristic polynomial. In particular, the characteristic polynomial is $r^3 - 1$. We solve this over $\mathbb{C}$ by completing the square: \begin{align*} 0 &= r^3 - 1 \\ &= (r - 1)(r^2 + r + 1) = (r - 1)\left(r^2 + r + \frac{1}{4} + \frac{3}{4}\right) \\ &= (r - 1)\left(\left(r + \frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}i\right)^2\right) \\ &= (r - 1)\left(r + \frac{1}{2} - i\frac{\sqrt{3}}{2}\right)\left(r + \frac{1}{2} + i\frac{\sqrt{3}}{2}\right), \end{align*} yielding $r = 1$ or $r = \frac{-1 \pm i\sqrt{3}}{2}$ (there are many ways to do this, including quadratic formula or $n$th roots of unity formulae, but I just like completing the square).

So we have a real root $r = 1$, which says that one fundamental solution will be $e^x$. We also have a pair of complex conjugates $r = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$, which suggests the other fundamental solutions are $e^{-x/2}\cos(x\sqrt{3}/2)$ and $e^{-x/2}\sin(x\sqrt{3}/2)$. Therefore, the general solution of the differential equation (ignoring initial values) is $$f(x) = Ae^x + Be^{-\frac{x}{2}}\cos\left(\frac{\sqrt{3}}{2} x\right) + Ce^{-\frac{x}{2}}\sin\left(\frac{\sqrt{3}}{2} x\right).$$ Differentiating, using product rule when necessary, \begin{align*} f'(x) &= Ae^x + \left(\frac{\sqrt{3}}{2}C - \frac{1}{2}B\right)e^{-\frac{x}{2}}\cos\left(\frac{\sqrt{3}}{2} x\right) + \left(-\frac{\sqrt{3}}{2}B - \frac{1}{2}C\right)e^{-\frac{x}{2}}\sin\left(\frac{\sqrt{3}}{2} x\right) \\ f''(x) &= Ae^x + \left(-\frac{\sqrt{3}}{2}C - \frac{1}{2}B\right)e^{-\frac{x}{2}}\cos\left(\frac{\sqrt{3}}{2} x\right) + \left(\frac{\sqrt{3}}{2}B - \frac{1}{2}C\right)e^{-\frac{x}{2}}\sin\left(\frac{\sqrt{3}}{2} x\right) \end{align*} Using the values of $f(0)$, $f'(0)$, and $f''(0)$, we can compute these constants. We have \begin{align*} 1 &= f(0) = A + B \\ 0 &= f'(0) = A - \frac{1}{2} B + \frac{\sqrt{3}}{2} C \\ 0 &= f''(0) = A - \frac{1}{2} B - \frac{\sqrt{3}}{2} C. \end{align*}

Adding all three equations yields $3A = 1$, hence $A = \frac{1}{3}$. Adding the second two equations yields $0 = 2A - B = \frac{2}{3} - B$, hence $B = \frac{2}{3}$. Subtracting the second two equations yields $C = 0$. So, our one and only solution is, $$f(x) = \frac{1}{3}e^x + \frac{2}{3}e^{-\frac{x}{2}}\cos\left(\frac{\sqrt{3}}{2} x\right).$$

Answer 2

Hint: start with Taylor's expansion for the exponential $$ e^{\omega x}=\sum_{k=0}^\infty\frac{\omega^k x^k}{k!} $$ and try to find $\omega\in\Bbb C$ such that $$ \langle\text{a polynomial of }\omega^k\rangle=\begin{cases} 0 & \text{if }3\nmid k,\\ \ne 0 & \text{if }3\mid k. \end{cases} $$ Play around with the third root of unity and the formula $$ \alpha^3-1=(\alpha-1)(\alpha^2+\alpha+1). $$