Find $\lim_{n \to \infty}a_n$ where $a_1=1$ and $a_{n+1}=a_n+\frac{1}{2^na_n}$. There are some attempts as follows.
Obviously, $\{a_n\}$ is increasing. Thus $a_n\ge a_1=1$. Therefore
$$a_{n+1}-a_n=\frac{1}{2^na_n}\le \frac{1}{2^n},$$
which gives that
$$a_n=a_1+\sum_{k=1}^{n-1}(a_{k+1}-a_k)\le a_1+\sum_{k=1}^{n-1}\frac{1}{2^k}=2-\frac{1}{2^{n-1}}.$$
Hence, $a_n\le 2$. Similarily,
$$a_{n+1}-a_n=\frac{1}{2^na_n}\ge \frac{1}{2^{n+1}},$$
which gives that
$$a_n=a_1+\sum_{k=1}^{n-1}(a_{k+1}-a_k)\ge a_1+\sum_{k=1}^{n-1}\frac{1}{2^{k+1}}=\frac{3}{2}-\frac{1}{2^{n}}.$$
Put the both aspects together, we have
$$\frac{3}{2}-\frac{1}{2^n}\le a_n\le 2-\frac{1}{2^{n-1}}.$$
But this does not satisfy the applying conditions of the squeeze theorem. Any other solutions?

When written under this form : $\quad 2^na_{n+1}a_n=2^n{a_n}^2+1$
We see that we can get rid of the $2^n$ term by setting $\ b_n=\sqrt{2^n}\,a_n\implies \frac 1{\sqrt{2}}b_{n+1}b_n={b_n}^2+1$
And we reduced the study to $$b_{n+1}=\sqrt{2}\left(b_n+\frac 1{b_n}\right)$$
Which according to this question Study convergence of $x_{n+1} = a\left(x_n + {1\over x_n}\right)$, for $x_1 = a$ and $a \in (0, 1)$ is divergent since $\sqrt{2}\ge 1$, which was expected provided $a_n$ has a limit (it is cauchy from $a_{n+1}-a_n<2^{-n}$ Proving a Sequence Converges - Cauchy?).
Maybe you can try your way in finding an asymptotic developpement for $b_n$ from there; but I fear a closed form would be difficult, despite its apparent simplicity it is uncooperative...
