1. ## double sequence

Let $d$ be a real number. For each integer $m \geq 0$, define a sequence $\{a_{m}(j) \}$, $j = 0,1,2, \ldots$ by the condition $a_{m}(0) = d/2^{m}$, and $a_{m}(j+1) = (a_{m}(j))^{2}+2a_{m}(j), \ j \geq 0$. Find $\lim_{n \to \infty} a_{n}(n)$.

Basically I found that the sequence $a_{n}(n)$ monotone increasing. But the offsets are becoming smaller and smaller. It looks like a taylor series.

Any ideas about how to get the limit?

2. Hello,
Originally Posted by particlejohn
Let $d$ be a real number. For each integer $m \geq 0$, define a sequence $\{a_{m}(j) \}$, $j = 0,1,2, \ldots$ by the condition $a_{m}(0) = d/2^{m}$, and $a_{m}(j+1) = (a_{m}(j))^{2}+2a_{m}(j), \ j \geq 0$. Find $\lim_{n \to \infty} a_{n}(n)$.

Basically I found that the sequence $a_{n}(n)$ monotone increasing. But the offsets are becoming smaller and smaller. It looks like a taylor series.

Any ideas about how to get the limit?
While playing with the first terms of the sequence I noticed that $a_n(n)=\left( a_n(0)+1\right)^{2^n}-1$ is true for $n\leq2$. Using induction it can be shown that it is in fact true for any non-negative integer $n$.

Induction hypothesis $\mathcal{P}_n\,:$ for all non-negative integer $p$, $\,a_p(n)=\left(a_p(0)+1\right)^{2^n}-1$

Let's show $\mathcal{P}_0$ : let $p$ be a non-negative integer. One has $a_p(0)=a_p(0)+1-1=(a_p(0)+1)^{2^0}-1$ so $\mathcal{P}_0$ is true.

Let's assume that there exists a non-negative integer $n$ such that $\mathcal{P}_n$ is true. Let $p$ be a non-negative integer.

\begin{aligned}
a_p(n+1)&=a_p(n)^2+2a_p(n)\\
&=\left[a_p(n)+1\right]^2-1\\
&=\left[\left(a_p(0)+1\right)^{2^n}-1+1\right]^2-1 \text{ since } a_p(n)=\left(a_p(0)+1\right)^{2^n}-1\\
&=\left[a_p(0)+1 \right]^{2^{n+1}}-1.\end{aligned}

This shows $\mathcal{P}_{n+1}$ and we have, for any non-negative integers $n$ and $p$, $\,a_p(n)=\left(a_p(0)+1\right)^{2^n}-1$. In particular, for $n=p$ : $a_n(n)=\left( a_n(0)+1\right)^{2^n}-1=\left( \frac{d}{2^n}+1\right)^{2^n}-1$. Now, what is $\lim_{n\to\infty}\left( 1+\frac{d}{2^n}\right)^{2^n}-1$ ?

3. $e^{d}-1$.

4. Originally Posted by particlejohn
$e^{d}-1$.
It seems good to me, I got the same result.

5. basically $\lim_{n\to\infty}\left( 1+\frac{d}{2^n}\right)^{2^n}-1$ behaves as $\lim_{n \to \infty} \left(1+\frac{d}{n} \right)^{n} -1$ right? Hence we get $e^{d}-1$.

6. Originally Posted by particlejohn
basically $\lim_{n\to\infty}\left( 1+\frac{d}{2^n}\right)^{2^n}-1$ behaves as $\lim_{n \to \infty} \left(1+\frac{d}{n} \right)^{n} -1$ right?
Yes.

7. Originally Posted by particlejohn
basically $\lim_{n\to\infty}\left( 1+\frac{d}{2^n}\right)^{2^n}-1$ behaves as $\lim_{n \to \infty} \left(1+\frac{d}{n} \right)^{n} -1$ right? Hence we get $e^{d}-1$.
Yes. This is a subsequence of $\lim_{n \to \infty} \left(1+\frac{d}{n} \right)^{n} -1$. Since this sequence converges to $e^d - 1$. Thus, any subsequence converges to $e^d - 1$.