# 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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