Example 3 Given $a_{1}=5, a_{i}(i=1,2, \cdots)$ are all natural numbers, satisfying the conditions $a_{1}<a_{2}<\cdots<a_{n}<\cdots$, and $a_{i+a}=2 a_{i}(i=1,2, \cdots)$, try to find $a_{1993}$.
Solve: First, calculate a few special values: $a_{1}=1+4, a_{6}=a_{1+a_{1}}=2 a_{1}=10=6+4, a_{16}=a_{6+a_{1}}=20=16+$ $4, \cdots$, from which we conjecture that $a_{n}=n+4$. We will prove this by mathematical induction.
When $n=1$, it is obviously true. Assume $a_{n}=n+4$, then $a_{2 n+4}=2 a_{n}=2(n+4)=(2 n+4)+4$.
Since $n+4=a_{n}<a_{n+1}<a_{n+2}<\cdots<a_{2 n+4}=(2 n+4)+4$, it follows that $a_{n+1}=(n+1)+4$.
Thus, $a_{n}=n+4$ holds for all natural numbers $n$.
Therefore, $a_{1993}=1993+4=1997$.
/Comment: Here, the conclusion is derived using inequalities and the properties of natural numbers, the approach is ingenious and thought-provoking.