Evaluating a series with some given formula

Question

I have a formula for the power series corresponding to the function $$\frac{z^{3k}}{(3k)!}$$ and I need to evaluate a new series with it but I can't see how to manipulate it even though I've had some ideas... Here it is : the series $$\frac{27^k}{(3k+1)!}.$$ I've tried to change it into $$\frac{3^{3k}}{(3k+1)!}$$ and then make the exponent of 3 3k+1 rather than 3k but it gives an expression that I can't use. I've had the idea of writing equalities with sums for multiples of 3, multiples of 3 substracting 1 and adding 1,... but it doesn't seem efficient... Any hint would be of great help, thank you!

Answer 1

The seems to follow the pattern of $\sin$. For $k \ge 1$, $$ \frac{d^{3}}{dx^{3}}\frac{x^{3k}}{(3k)!} = \frac{x^{3k-3}(3k)(3k-1)(3k-2)}{3k!} = \frac{x^{3(k-1)}}{(3(k-1)!)}. $$ Let $$ f(x)=\sum_{k=0}^{\infty}\frac{x^{3k}}{(3k)!}. $$ Then $f'''(x) = f$ with $f(0)=1$, $f'(0)=0$, $f''(0)=0$. The solution of this ODE comes directly from a factoring $$ (D^{3}-1)=(D-1)(D-\alpha)(D-\alpha^{2}),\;\;\; \alpha=e^{\frac{2\pi}{3}i} $$ $\alpha = \cos(2\pi/3)+i\sin(2\pi/3)=-1/2+i\sqrt{3}/2$. The solution is a linear combination $$ f(x) = Ae^{x}+Be^{\alpha x}+Ce^{\alpha^{2} x}. $$ Try $A=B=C=1/3$. Then $f(0)=0$. The sum of the cube roots of 1 is $1+\alpha+\alpha^{2}$, which is known to be 0. You can see graphically, or you can show algebraically using $\alpha^{3}=1$ that $(1+\alpha+\alpha^{2})(1-\alpha)=0$ implies the result because $1-\alpha \ne 0$. Therefore, $$ \begin{align} f'(0) & = \frac{1}{3}(1+\alpha+\alpha^{2})=0 \\ f''(0) & = \frac{1}{3}(1+\alpha^{2}+\alpha^{4})=(1+\alpha^{2}+\alpha)=0. \end{align} $$ In terms of real functions $$ f(x) = \frac{1}{3}e^{x}+\frac{2}{3}e^{-x/2}\cos(\sqrt{3}x/2). $$