I was reading about triangular numbers on Wikipedia, where the following statement is made:
The number of line segments between closest pairs of dots in the triangle can be represented in terms of the number of dots or with a recurrence relation: $$L_n=3T_{n-1}=3\binom{n}2{}; $$ $$L_n=L_{n-1}+3(n-1),L_1=0$$ In the limit, the ratio between the two numbers, dots and line segments is $$lim_{n \rightarrow \infty}\frac{T_n}{L_n}=\frac{1}{3}$$
I understand how we get $L_n=L_{n-1}+3(n-1),L_1=0$: for every new side we add, we add $3$ line segments for every triangle that can be formed from that new side, and this comes out to $3(n-1)$ more segments for each increment of $L_n$.
However, I don't understand how $$L_n=3T_{n-1}=3\binom{n}2{} $$ and $$lim_{n \rightarrow \infty}\frac{T_n}{L_n}=\frac{1}{3}$$ were derived.
For $L_n = 3T_{n-1}$, visualize all the small line segments being drawn. Then you will have many small equilateral triangles, $T_{n-1}$ of which point "upward." Each of the small line segments is an edge of exactly one of these "upward" small equilateral triangles, so there are $3 T_{n-1}$ segments.
For the limit, $$\frac{T_n}{L_n} = \frac{T_n}{3 T_{n-1}} = \frac{1}{3} \frac{n(n-1)}{(n-1)(n-2)} \to \frac{1}{3}.$$