Help With finding Taylor Series to evaluate the sum $\sum_{n=0}^{\infty}\frac{1}{(3n)!}$

Question

I have tried finding a Taylor series to evaluate the sum $\sum_{n=0}^{\infty}\frac{1}{(3n)!}$, but am unable to find such a function. Is there a Taylor Series that could frame this sum?

Answer 1

Hint: Find a differential equation for $$ f(x) = \sum_{n=0}^\infty \frac{x^{3n}}{(3n)!} $$ Differentiate this 3 times and relate it to the original.

Answer 2

If one has a function given by a convergent power series $$f(x)=\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+\cdots$$ then the sum of every third term $$g(x)=\sum_{n=0}^\infty a_{3n}x^{3n}=a_0+a_3x^3+a_6x^6+a_9x^9+\cdots$$ may be obtained by the trick of "series trisection".

Write $\newcommand{\om}{\omega}\om=\exp(2\pi i/3)=\frac12(-1+i\sqrt3)$, so that $\om^3=1$, and $1+\om+\om^2=0$. Then $$f(\om x)=a_0+\om a_1x+\om^2 a_2x^2+a_3x^3+\cdots$$ and $$f(\om^2 x)=a_0+\om^2 a_1x+\om a_2x^2+a_3x^3+\cdots.$$ Adding these to each other, and to $f(x)$ gives $$f(x)+f(\omega x)+f(\omega^2x)=3a_0+3a_3x^3+3a_6x^6+\cdots=3g(x).$$

In your application, you want to compute $g(1)$ where $f(x)=\exp(x)$.