How can we prove that? $$\sum_{n=1}^\infty\frac{x^{3n}}{(3n-1)!}=\frac{1}{3}e^{\frac{-x}{2}}x\left(e^{\frac{3x}{2}}-2\sin\left(\frac{\pi+3\sqrt{3}x}{6}\right)\right).$$

I think if we write the taylor expansion of $\sin(u)$ and $e^u$, we can arrive from RHS to LHS, but I am looking for a way to prove it from LHS.

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    $\begingroup$ Not my area of expertise. I could probably devise a solution myself through differential equations, but I imagine that's off the table. $\endgroup$
    – Mike
    Dec 17, 2013 at 15:50
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    $\begingroup$ This is a general trick to extract a (regular) sub-series from a series : the multiplication formula. For example to obtain the expansion of $\zeta$ around $0$ keeping only every third term you could do series(zeta(x)+zeta(xexp(-2*PII/3))+zeta(xexp(2*PII/3)),x)/3 (add three times the function taken at x multiplied by a root of unity). $\endgroup$ Dec 17, 2013 at 15:50
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    $\begingroup$ (in your case the function zeta should be replaced by $\;\displaystyle x\mapsto x\,e^x$) $\endgroup$ Dec 17, 2013 at 16:47
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    $\begingroup$ @Did: What thing is strange in my question that you want to close my question? $\endgroup$
    – user91500
    Dec 20, 2013 at 17:46
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    $\begingroup$ Would you describe "How can we obtain following identity? [Identity]" and nothing else, as the ideal MSE question? $\endgroup$
    – Did
    Dec 20, 2013 at 18:03

4 Answers 4


Consider the third root of unity $\rho = e^{2\pi i/3} = \frac{-1+i\sqrt{3}}{2}$. You have

$$e^{\rho z} = \sum_{k=0}^\infty \frac{\rho^k z^k}{k!} = \sum_{m=0}^\infty \frac{z^{3m}}{(3m)!} + \rho\sum_{m=0}^\infty \frac{z^{3m+1}}{(3m+1)!} + \rho^2\sum_{m=0}^\infty \frac{z^{3m+2}}{(3m+2)!}$$

since $\rho^{3m} = 1,\; \rho^{3m+1} = \rho,\; \rho^{3m+2} = \rho^2$. You have something similar for $e^{\rho^2 z}$. Also consider $1 + \rho + \rho^2 = 0$. Then a suitable combination of $e^{\rho^k x}$ gives you

$$\sum_{n=1}^\infty \frac{x^{3n-1}}{(3n-1)!}.$$

Using Euler's formula $e^{it} = \cos t + i\sin t$ then gives you the right hand side.

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    $\begingroup$ Thanks very much. I found suitable combination:$$\sum_{m=0}^\infty\frac{z^{3m+2}}{(3m+2)!}=\frac{\rho e^{\rho^2z}-\rho e^z+e^{\rho z}-e^z}{2\rho^2-\rho-1}.$$ $\endgroup$
    – user91500
    Dec 18, 2013 at 7:53

Too long for a comment: Just to put things into proper perspective:

$$\sum_{n=0}^\infty\frac{x^{an+b}}{(an+b)!}= \begin{cases} e^x,&(a,b)=(1,0)\\ \\ \cosh x,&(a,b)=(2,0)\\ \\ \sinh x,&(a,b)=(2,1)\\ \\ \dfrac13\bigg[e^x+2e^{^{-\tfrac x2}}\cos\bigg(x\dfrac{\sqrt3}2\bigg)\bigg],&(a,b)=(3,0)\\ \\ \dfrac13\bigg\{e^x-e^{^{-\tfrac x2}}\bigg[\cos\bigg(x\dfrac{\sqrt3}2\bigg)-\sqrt3\sin\bigg(x\dfrac{\sqrt3}2\bigg)\bigg]\bigg\},&(a,b)=(3,1)\\ \\ \dfrac13\bigg\{e^x-e^{^{-\tfrac x2}}\bigg[\cos\bigg(x\dfrac{\sqrt3}2\bigg)+\sqrt3\sin\bigg(x\dfrac{\sqrt3}2\bigg)\bigg]\bigg\},&(a,b)=(3,2)\\ \\ \tfrac12(\cosh x+\cos x),&(a,b)=(4,0)\\ \\ \tfrac12(\sinh x+\sin x),&(a,b)=(4,1)\\ \\ \tfrac12(\cosh x-\cos x),&(a,b)=(4,2)\\ \\ \tfrac12(\sinh x-\sin x),&(a,b)=(4,3) \end{cases}$$

  • $\begingroup$ Also, for $(a,b)=\bigg(1, \dfrac12\bigg),~$ we have $S=e^x~\text{erf} \big(\sqrt x\big).$ $\endgroup$
    – Lucian
    May 15, 2015 at 19:56
  • $\begingroup$ Also, for $(a,b)=\bigg(\dfrac12, 0\bigg),~$ starting from $n=0,$ as well as for $(a,b)=\bigg(\dfrac12, \dfrac12\bigg),~$ starting from $n=-1,$ we have $S=e^x~\Big[1+\text{erf} \big(\sqrt x\big)\Big].$ $\endgroup$
    – Lucian
    Jan 25, 2017 at 13:56

Consider the function


Consider this a Maclaurin series for $f(x)$. So we have $f(0)=0,f'(0)=0$ and $f''(0)=1$. Finally, take note that $f'''(x)=f(x)$. Solve this initial value problem and multiply by $x$ to get your answer.

It looks like it may take some more work to get this into the desired form, however, so I'll continue. The characteristic equation for our problem is $s^3-1=0$, whose roots are the third roots of unity $1$ and $-\frac12\pm\frac{\sqrt3}2i$. This suggests

$$f(x)=k_1e^x+k_2e^{-\frac x2}\sin(\frac{x\sqrt3}2)+k_3e^{-\frac x2}\cos(\frac{x\sqrt3}2)=$$ $$k_1e^x+e^{-\frac x2}[k_2\sin(\frac{x\sqrt3}2)+k_3\cos(\frac{x\sqrt3}2)]$$ $$f'(x)=k_1e^x+e^{-\frac x2}[(-\frac12k_2-\frac{\sqrt3}2k_3)\sin(\frac{x\sqrt3}2)+(-\frac12k_3+\frac{\sqrt3}2k_2)\cos(\frac{x\sqrt3}2)]$$ $$f''(x)=k_1e^x+e^{-\frac x2}[(\frac14k_2+\frac{\sqrt3}4k_3+\frac{\sqrt3}4k_3-\frac34k_2)\sin(\frac{x\sqrt3}2)+(\frac14k_3-\frac{\sqrt3}4k_2-\frac{\sqrt3}4k_2-\frac34k_3)\cos(\frac{x\sqrt3}2)]=$$ $$k_1e^x+e^{-\frac x2}[(-\frac12k_2+\frac{\sqrt3}2k_3)\sin(\frac{x\sqrt3}2)+(-\frac12k_3-\frac{\sqrt3}2k_2)\cos(\frac{x\sqrt3}2)]$$

Plugging in initial conditions

$$k_1+k_3=0$$ $$k_1+\frac{\sqrt3}2k_2-\frac12k_3=0$$ $$k_1-\frac{\sqrt3}2k_2-\frac12k_3=1$$

From the first equation, we get $k_3=-k_1$. Adding the second and third gives $k_3=2k_1-1$. So $k_1=\frac13,k_3=-\frac13$ and

$$\frac13+\frac{\sqrt3}2k_2+\frac16=0$$ $$\frac{\sqrt3}2k_2=-\frac12$$ $$k_2=-\frac{\sqrt3}3$$

So the last step is to prove


This last step can be done by solving $a\cos b=\sqrt3$ and $a\sin b=1$. This gives $\tan b=\frac{\sqrt3}3$ and $a^2=3+1=4$.


This is actually calling for Laplace transform. Set $f(x)=\sum_{n=1}^\infty \frac{x^{3n}}{(3n-1)!}$ then \begin{align} F(s) &= \int_0^\infty \frac{f(x)}{x} \, e^{-sx} \, {\rm d}x \\ &= \sum_{n=1}^\infty \frac{1}{s^{3n}} \int_0^\infty \frac{u^{3n-1} \, e^{-u}}{(3n-1)!} \, {\rm d}u \\ &= \frac{1}{s^3-1} \end{align} Now \begin{align} \frac{f(x)}{x} = \frac{1}{2\pi i} \int_{\gamma -i\infty}^{\gamma + i\infty} F(s) \, e^{xs} \, {\rm d}s \end{align} where $\gamma > 1$. The contour can be closed along the half-circle to the left and using the residue-theorem you will get \begin{align} \frac{f(x)}{x} &= \frac{e^{-\frac{x}{2}}}{3} \left( e^{\frac{3x}{2}} + e^{\frac{i\sqrt{3}x}{2}+\frac{2\pi i}{3}} + e^{-\frac{i\sqrt{3}x}{2}-\frac{2\pi i}{3}} \right) \\ &= \frac{e^{-\frac{x}{2}}}{3} \left( e^{\frac{3x}{2}} + 2\cos\left(\frac{\sqrt{3}x}{2}+\frac{2\pi}{3}\right) \right) \end{align}


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