Simplify $\lim_{k\rightarrow 0} \sum_{n=1}^{\infty} \frac{(-1)^{n!+k!+n+2}}{k!(2n+1)!}$

Problem

Find the value of $S$. ¹

$$(★): S \equiv \lim_{k\rightarrow 0} \sum_{n=1}^{\infty} \frac{(-1)^{n!+k!+n+2}}{k!(2n+1)!}$$

Solution

$$\begin{align} S &= \lim_{k\rightarrow 0} \sum_{n=1}^{\infty} \frac{(-1)^{n!} (-1)^{k!} (-1)^n \cancelto{1}{(-1)^2}}{k!(2n+1)!}\\ &= \lim_{k\rightarrow 0} \frac{(-1)^{k!}}{k!} \sum_{n=1}^{\infty} \frac{(-1)^{n!} (-1)^n}{(2n+1)!} \end{align}$$ We make the following observations.

1. $n!$ is always even all natural numbers $n \geq 2$ because $$n! = n\cdot(n-1)\cdot(n-2)\cdot\ldots\cdot 2 \cdot 1.$$ Therefore, $(-1)^{n!} = 1$ for any natural number $n\geq2$.
2. The limit in (★) only makes sense if we define $k!$ and $-1$ as follows. $$\begin{aligned} k! &\equiv \Gamma(k+1)\\ -1 &\equiv e^{j\pi} \end{aligned}$$ Here, $\Gamma(k+1)$ is the gamma function and $j$ is the imaginary unit. We then get $$\lim_{k\rightarrow 0} \frac{(-1)^{k!}}{k!} = \lim_{k\rightarrow 0} e^{j\pi (k!)} = -1$$ because $$\lim_{k\rightarrow 0} {k!} = \lim_{k\rightarrow 0} {\Gamma(k+1)} = \Gamma(1) = 1.$$

We then obtain $$\begin{align} S &= -\left( \frac{(-1)^{1!}(-1)^1}{(2\cdot 1+1)!} + \sum_{n=2}^{\infty} \frac{(-1)^n}{(2n+1)!} \right)\\ &= -\left( \frac{1}{3!} + \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} -\frac{(-1)^0}{(2\cdot 0+1)!} -\frac{(-1)^1}{(2\cdot 1+1)!} \right)\\ &= \frac{2}{3} - \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}. \end{align}$$ We observe that $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} 1^n = \sin{1}$$ because for any real number $x$, the Taylor expansion of $\sin{x}$ is, $$\sin{x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^n.$$

Therefore,

$$S = \frac{2}{3} - \sin{1}.■$$

References

¹ Author Unknown. Problem taken from a Facebook post.