How can I compute the limit of the following series :$ \sum \frac{1}{(3k)!} $?
Asked
Active
Viewed 93 times
1
-
1What are your limits on the sum? – TomGrubb Jan 04 '16 at 22:42
-
4I would try considering $e^{x}+e^{ax}+e^{a^2x}$ for a suitable complex number $a$. – Jan 04 '16 at 22:44
-
@RossMillikan per the hint from user Normal, it converges to $e + e^\zeta + e^{\zeta^2}$ where $\zeta$ is a primitive cube root of 1. – hunter Jan 04 '16 at 22:51
-
1@hunter To one third of that. – Jan 04 '16 at 22:53
-
much thanks for the hint ! – kosskor Jan 04 '16 at 22:54
-
A related question. – Lucian Jan 05 '16 at 05:47
-
Look up multisection of series. Your problem is a multisection of $e^x$. – marty cohen Jan 06 '16 at 01:27
1 Answers
2
Consider decimation of the series $f(z) = \sum_{k \ge 0} a_k z^k$. Call $\omega = \mathrm{e}^{\frac{2 \pi \mathrm{i}}{m}}$, a primitive $n$-th root of 1. It is easy to check that:
$\begin{align} \sum_{0 \le k < m} \omega^{r k} = \begin{cases} m & m \mid r \\ 0 & \text{otherwise} \end{cases} \end{align}$
Thus:
$\begin{align} \sum_{0 \le r < m} f(z \omega^r) &= \sum_{0 \le r < m} \sum_{k \ge 0} a_k z^k \omega^{r k} \\ &= \sum_{k \ge 0} a_k z^k \sum_{0 \le r < m} \omega^{r k} \\ &= m \sum_{k \ge 0} a_{k m} z^{k m} \end{align}$
Your series is nothing but:
$\begin{align} \frac{e^{\omega x}}{3} \end{align}$
with $\omega = \mathrm{e}^{\frac{2 \pi \mathrm{i}}{3}}$
vonbrand
- 27,812