How to interpret these combinatorics equations' combination meanings

Question

Let$$S_n:=\sum_{i=0}^n\frac{(-1)^i}{2i+1}\binom ni\\ H_n:=\sum_{i=0}^n\frac{(-1)^i}{2i+3}\binom ni$$then$$\begin{align}S_{n+1}-S_n&=\sum_{i=1}^n\frac{(-1)^i}{2i+1}\binom{n}{i-1}+\frac{(-1)^{n+1}}{2n+3}\\&=-\sum_{i=0}^{n-1}\frac{(-1)^i}{2i+3}\binom{n}{i}+\frac{(-1)^{n+1}}{2n+3}\\&=-H_n&(1)\end{align}$$and$$\begin{align}S_n&=\sum_{i=0}^n\frac{(-1)^i}{2i+1}\binom ni\\ &=\sum_{i=0}^n(-1)^i\frac{2i+1-2i}{2i+1}\binom ni\\ &=\sum_{i=0}^n(-1)^i\binom ni-2n\sum_{i=1}^{n}\frac{(-1)^i}{2i+1}\binom{n-1}{i-1}\\ &=2n\sum_{i=1}^{n}\frac{(-1)^{i-1}}{2i+1}\binom{n-1}{i-1}\\ &=2n\sum_{i=0}^{n-1}\frac{(-1)^{i}}{2i+3}\binom{n-1}{i}\\ &=2nH_{n-1}&(2)\end{align}$$so that$$S_n=(2n+3)H_n\implies S_{n+1}=\frac{2n+2}{2n+3}S_n\tag3$$by induction$$\forall n\in\mathbb N,\ S_n=\frac{4^n(n!)^2}{(2n+1)!}\tag4$$I want to find some combination meanings to explain $S_n,H_n,(1),(2),(3),(4)$ but have no idea about it.