An interesting sequence and its limit

Problem: Consider a sequence of numbers \(a_m(j)\) such that \(a_m(j+1) = a_m(j)^2 + 2 a_m(j)\) for all \(j \geq 0\) and \(m \geq 0\) with \(a_m(0)=\frac{d}{2^m}\) for some real number \(d\). Determine \(\displaystyle\lim_{m \to +\infty}a_m(m)\).

Solution:  The above recursive relation can be rewritten as follows:

\(1+a_m(j+1)=(1+a_m(j))^2\)

One of the most underrated tricks in solving recusive relations is to substitute in a new sequence whose pattern has a clear path to solution.  In this case the substitution glares at you. Let \(b_m(j)=1+a_m(j)\).  The new sequence satisfies the recurrence

\(b_m(j+1)=b_m(j)^2\)

and has an initial condition \(b_m(0)=1+\frac{d}{2^m}\).  We then can step through the first few cases:

\(b_m(1)=(1+\frac{d}{2^m})^2\),

\(b_m(2)=(1+\frac{d}{2^m})^4\),

until we get to the desired case:

\(b_m(m)=(1+\frac{d}{2^m})^{2^m}\).

Using the known limiting relation \(\displaystyle\lim_{m \to +\infty}(1+\frac{z}{m})^m=\exp{z}\), and the relation \(b_m(j)=1+a_m(j)\), we get the desired result:

\(\displaystyle\lim_{m \to +\infty}a_m(m)=\exp{d}-1\).

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Ron Gordon

Math nerd in his early 40's who seems to have an opinion about everything and an inability to keep it to himself.