Shoulders, Giants, and Cats🐈

Problem Find the value of $S$. $$ (★): S \equiv \lim_{k\rightarrow 0} \sum_{n=1}^{\infty} \frac{(-1)^{n!+k!+n+2}}{k!(2n+1)!} $$ 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. $n!$ 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 $n\geq2$. 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_

Problem Find the limit $\sin(\sin(...\sin(x)))$ for any real number $x$. Solution The problem can be precisely formulated as follows: Find the limit $L$ such that $$ L \equiv\lim_{n\rightarrow\infty}{f_n(x)}, $$ where $x\in\mathbb{R}$, $n\in\mathbb{N}$, and $$\begin{align} f_1(x) &= \sin(x)\\ f_n(x) &= \sin(f_{n-1}(x)). \end{align}$$ We observe that, if the limit $L$ exists, $$\begin{align} \lim_{n\rightarrow\infty} f_{n-1}(x) &= L\\ \lim_{n\rightarrow\infty} f_n(x) &= L, \end{align}$$ and thus, $$ L = \sin{L}, $$ which we show to be $L=0$ in the following. First, we show that the limit exists. We accept the Axiom of Continuity of the Real Line, which says that any bounded monotonic sequence converges. Thus, to show that $L$ exists, we need only show that (1) $f_n(x)$ is a bounded sequence of $n$, and that (2) $f_n(x)$ is a monotonic sequence of $n$. $f_n(x)$ is clearly bounded because $f_n(