complex series exponential evaluation

Question

How do we evaluate the series $$\sum \frac{27^n}{(3n+1)!}$$ I could simplify this to $$\sum \frac{3^{3n}}{(3n+1)!}$$ but typically the tables provide you with the general series form of $$\sum \frac{x^{an+b}}{(an+b)!}$$ How do I go from herer? The b's are not the same!

Answer 1

You can write the series as $$\frac{1}{3}\sum \frac{3^{3n+1}}{(3n+1)!}$$right?

Answer 2

HINT:

As $1+\omega+\omega^2=0$ where $\omega$ is a complex root of unity,

$$e^x=\sum_{r=0}^\infty\frac{x^r}{r!}$$

$$e^{\omega x}=\cdots$$

$$e^{\omega^2 x}=\cdots$$

$$e^x+\omega^2\cdot e^{\omega x}+\omega\cdot e^{\omega^2 x}=?$$

Answer 3

Apply ratio test: $$\limsup_{n\to\infty}\left|\frac{27}{(3n+4)(3n+3)(3n+2)}\right|<1$$ So series converges by ratio test.