Closed form of a series

Question

I am looking for a closed form of the following convergent series: $$\sum_{n=0}^\infty \frac{(-\lambda^2)^n}{(6n+i)!}$$ For the case of $i=0$, the answer is ready, but when $i=1,2,3,4,5$, everything gets crazy.

Answer 1

The typical way to "skip terms" in a power series is to use roots of unity. Here, let $\zeta=e^{i\pi/3}$ be a primitive $6$th root of unity, so that $$ \frac16 \sum_{j=0}^5 (\zeta^\ell)^j = \begin{cases} 1, &\text{if } 6\mid \ell, \\ 0, &\text{otherwise.} \end{cases} $$ Therefore, setting $\alpha=e^{i\pi/6}$, for any $0\le k\le 5$, \begin{align*} \sum_{n=0}^\infty \frac{(-\lambda^2)^n}{(6n+k)!} &= (\alpha\lambda^{1/3})^{-k} \sum_{n=0}^\infty \frac{(\alpha\lambda^{1/3})^{6n+k}}{(6n+k)!} \\ &= (\alpha\lambda^{1/3})^{-k} \sum_{m=0}^\infty \frac{(\alpha\lambda^{1/3})^m}{m!} \begin{cases} 1, &\text{if } 6\mid (m-k) \\ 0, &\text{otherwise} \end{cases} \Big\} \\ &= (\alpha\lambda^{1/3})^{-k} \sum_{m=0}^\infty \frac{(\alpha\lambda^{1/3})^m}{m!} \frac16 \sum_{j=0}^5 (\zeta^{m-k})^j \\ &= \frac16 (\alpha\lambda^{1/3})^{-k} \sum_{j=0}^5 \zeta^{-jk} \sum_{m=0}^\infty \frac{(\alpha\lambda^{1/3}\zeta^j)^m}{m!} = \frac16 (\alpha\lambda^{1/3})^{-k} \sum_{j=0}^5 \zeta^{-jk} \exp(\alpha\lambda^{1/3}\zeta^j). \end{align*}