Closed form for the series $\sum\limits_{n=0}^\infty \frac{\exp(\cos(n))}{n!}$

Question

Does anyone have any ideas on how to find a closed form for the following expression? It comes up when trying to bound a particular integral. The sum is:

$$\sum_{n=0}^{\infty} \frac{e^{\cos(n)}}{n!}$$

Thank you very much for your thoughts.

Answer 1

I highly doubt there is any reasonable closed form expression. However, since you are just trying to bound some integral, this may be useful.

Denote the partial sums by $\displaystyle S_{m}=\sum_{n=0}^{m} \frac{e^{\cos(n)}}{n!}$. Note that $\displaystyle S_{\infty}=S_{m}+\sum_{m+1}^{\infty} \frac{e^{\cos(n)}}{n!}$. We can bound the error term above and below as follows:

$$\sum_{m+1}^{\infty} \frac{e^{\cos(n)}}{n!} \ge \frac{1}{e}\sum_{m+1}^{\infty} \frac{1}{n!}=1-\frac{\Gamma(m+1,1)}{\Gamma(m+1)}=E_{m}$$ $$\sum_{m+1}^{\infty} \frac{e^{\cos(n)}}{n!} \le e\sum_{m+1}^{\infty} \frac{1}{n!}=e^{2}E_{m}$$

Since we are summing positive terms, we may bound the final sum as follows:

$$S_{m}+E_{m} \le S_{\infty} \le S_{m}+e^{2}E_{m}$$

The majority of the sum is deposited in the first few terms, so plugging in $m=5$ yields the following decent, rigorous bounds:

$$2.14158 \le S_{\infty} \le 2.14538$$

Numerically, $S_{\infty} \approx 2.14506$.