Find the limit of the sequence $a_{n+1}=a_n + \frac{1}{2^n a_n}$ with $a_1=1$ Find the limit of the sequence $a_{n+1}=a_n + \frac{1}{2^n a_n}$ with $a_1=1$.
I could only estimate the upper and lower bound for $a_n$. My result is
$$\sqrt{3} \leq \lim_{n\to \infty} a_n \leq \sqrt{\frac{79}{24}}$$.
Is there any way to compute $\lim_{n\to \infty} a_n$, or find the non-linear equality it satisfies.
B.T.W., I find the limit value depends on the initial value $a_1=1$.

Write $L = \lim_{n \to \infty} a_n$. Squaring gives
$$a_{n+1}^2 = a_n^2 + \frac{1}{2^{n-1}} + \frac{1}{2^{2n} a_n^2}$$
which gives, ignoring the last term
$$L^2 \ge 1 + \sum_{k=0}^{\infty} \frac{1}{2^k} = 3$$
which I presume is where your lower bound of $\sqrt{3}$ comes from. Since $a_n$ is monotonically increasing, we have $a_n^2 \ge 1$ and so $\frac{1}{a_n^2} \le 1$, which gives
$$L^2 \le 3 + \sum_{k=1}^{\infty} \frac{1}{2^{2k}} = 3 + \frac{1}{3} = \frac{10}{3}$$
which gets us an upper bound of $\sqrt{ \frac{10}{3} } = \sqrt{ \frac{80}{24} }$, slightly worse than the one you obtained. In particular $a_n$ is both monotonic and bounded so the limit actually exists. By monotonicity this is an upper bound on the entire series, so we have $a_n^2 \le \frac{10}{3}$, which gives a stronger lower bound
$$L^2 \ge 3 + \sum_{k=1}^{\infty} \frac{1}{2^{2k} \frac{10}{3}} = 3 + \frac{1}{10}.$$
We can improve the lower and upper bounds starting from the exact value of any particular term $a_k$ of the sequence, as follows. Starting the recurrence from this term gives
$$L^2 \ge a_k^2 + \sum_{i=k-1}^{\infty} \frac{1}{2^i} = a_k^2 + \frac{1}{2^{k-2}}$$
as a lower bound, and since $a_n$ is monotonically increasing we have $a_n \ge a_k$ for $n \ge k$ so we get
$$L^2 \le a_k^2 + \frac{1}{2^{k-2}} + \sum_{i=k} \frac{1}{2^{2i} a_k^2} = a_k^2 + \frac{1}{2^{k-2}} + \frac{1}{3 \cdot 4^{k-1} a_k^2}.$$
As above this upper bound lets us strengthen the lower bound again, to
$$L^2 \ge a_k^2 + \frac{1}{2^{k-2}} + \frac{1}{3 \cdot 4^{k-1} \left( a_k^2 + \frac{1}{2^{k-2}} \right)}.$$
The first few terms are $a_1 = 1, a_2 = \frac{3}{2}, a_3 = \frac{5}{3}, a_4 = \frac{209}{120}$. Applying the above bounds to $a_3 = \frac{5}{3}$ to keep the arithmetic simple gives
$$\frac{25}{9} + \frac{1}{2} + \frac{3}{424} \le L^2 \le \frac{25}{9} + \frac{1}{2} + \frac{3}{400}$$
which gives $L \in [1.81242 \dots, 1.81253 \dots]$, and using larger terms would get us exponentially closer to $L$, but I doubt there's anything nice to say as far as closed forms.
