What is the sum of this series: $\displaystyle\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$

Question

I tried getting it into a closed form but failed. Could someone help me out?

$$\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$$

Answer 1

The function $f$ defined by $f(x)=\sum_{n=0}^\infty\frac{x^{3n}}{(3n)!}$ satisfies the differential equation $f^{(3)}(x)=f(x)$. So we can just set out to solve this differential equation with its initial conditions that $f(0)=1$, and $f'(0)=f''(0)=0$.

This homogeneous equation has characteristic polynomial $X^3-1$, with roots $1$, $\frac{-1+\sqrt{3}i}{2}$, and $\frac{-1-\sqrt{3}i}{2}$, and it follows that $$f(x)=Ae^x+Be^{\frac{-1+\sqrt{3}i}{2}x}+Ce^{\frac{-1-\sqrt{3}i}{2}x}$$

Use initial conditions and linear algebra to solve for $A$, $B$, and $C$:

\begin{align} f(0)&=1 & A+B+C&=1\\ f'(0)&=0& A+\frac{-1+\sqrt{3}i}{2}B+\frac{-1-\sqrt{3}i}{2}C&=0\\ f''(0)&=0& A+\frac{-1-\sqrt{3}i}{2}B+\frac{-1+\sqrt{3}i}{2}C&=0\\ \end{align}

Subtracting the last two equations reveals that $B=C$, and then the first and second equations reduce to: \begin{align} A+2B&=1\\ A-B&=0 \end{align} And now it is clear that $A=B=C=\frac13$. Then manipulate the last two terms to express them using cosine. So \begin{align} f(x)&=\frac13e^x+\frac13e^{\frac{-1+\sqrt{3}i}{2}x}+\frac13e^{\frac{-1-\sqrt{3}i}{2}x}\\ &=\frac13e^x+\frac13e^{-x/2}\left(e^{\frac{\sqrt{3}i}{2}x}+e^{\frac{-\sqrt{3}i}{2}x}\right)\\ &=\frac13e^x+\frac23e^{-x/2}\cos\left(\frac{\sqrt{3}}{2}x\right)\\\end{align}

Lastly, swap out $x$ for $\sqrt[3]{a}$:

$$\sum_{n=0}^\infty\frac{a^n}{(3n)!}=\frac13e^{\sqrt[3]{a}}+\frac2{3\sqrt{e}^{\sqrt[3]{a}}}\cos\left(\frac{\sqrt{3}}{2}\sqrt[3]{a}\right)$$

Answer 2

Here is an approach based on the Laplace transform techniques. Recalling the Laplace transform of a function $f$,

$$ F(s) = \int_{0}^{\infty} f(x) e^{-sx} dx \,, $$

and using the fact that the Laplace transform of $x^m$ is given by $\frac{\Gamma(m+1)}{x^{m+1}}$, we can compute the Laplace transform of $f(x)=\sum\limits_{n=0}^{+\infty}\frac{x^{3n}}{(3n)!}$ as

$$ F(s) = \sum_{k=0}^{\infty} \frac{1}{s^{3k+1}}=\frac{s^2}{s^3-1}$$

$$ \implies F(s) = \frac{1}{3\, \left( s-1 \right) }+\frac{1}{3\, \left( s+\frac{1}{2}-\frac{1}{2}\,i\sqrt {3}\right)}+\frac{1}{3\, \left( s+\frac{1}{2}+\frac{1}{2}\,i\sqrt {3} \right)}\longrightarrow (1)\,. $$

Taking the inverse Laplace transform of $(1)$ yields ,

$$f(x)= \frac{1}{3}e^x + \frac{1}{3}e^{\left(-\frac{1}{2}+\frac{1}{2}\,i\sqrt {3}\right)x}+\frac{1}{3}e^{\left(-\frac{1}{2}-\frac{1}{2}\,i\sqrt {3}\right)x} $$

$$\implies f(x)= \frac{1}{3}\,{{\rm e}^{x}}+\frac{2}{3}\,{{\rm e}^{-x/2}}\cos \left( \sqrt {3}x/2\right) \longrightarrow (*). $$

The last step is to substitute $x=a^{1/3}$ in $(*)$.

Note: The Laplace transform of $e^{ax}$ is given by

$$ \mathcal{L}(e^{ax})(s) = \frac{1}{s-a} $$

Answer 3

Hint: Your sum equals $$ \sum_{m=0}^\infty \frac{(a^{1/3})^m}{m!} \frac{1 + e^{2\pi i m/3} + e^{-2\pi i m/3}}3. $$ You should be able to split this up into three instances of the power series for $e^y$. That should allow you to see how Maple got the answer it did.

Answer 4

You should know the Taylor series for $e^x$:

$e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \cdots$

Then, the Taylor series for $e^{-x}$ is:

$e^{-x} = 1 + (-x) + \dfrac{(-x)^2}{2} + \dfrac{(-x)^3}{3!} + \dfrac{(-x)^4}{4!} + \cdots$

$e^{-x} = 1 - x + \dfrac{x^2}{2} - \dfrac{x^3}{3!} + \dfrac{x^4}{4!} - \cdots$

Add these two series together, and all the odd terms cancel. So you get:

$e^x+e^{-x} = 2 + 2\dfrac{x^2}{2} + 2\dfrac{x^4}{4!} + \cdots$

$\dfrac{e^x+e^{-x}}{2} = 1 + \dfrac{x^2}{2} + \dfrac{x^4}{4!} + \cdots$

Now, plug in $x = \sqrt{a}$, and see what you get.