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?
 A: 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...
A: Here is a short proof that the limit is between 1.810 and 1.828. This can be improved to get better precision, I believe.
Squaring both sides of the recurrence, we get
$$
a_{n+1}^2 = a_n^2 + \frac{2}{2^n} + \frac{1}{4^n a_n^2} \tag{1}
$$
and so, summing,
$$
a_n^2 = a_1^2 + \sum_{k=1}^{n-1} (a_{k+1}^2 - a_k^2) = 1+\sum_{k=1}^{n-1}\frac{2}{2^k} + \sum_{k=1}^{n-1}\frac{1}{4^k a_k^2} \tag{2}
$$
and so the limit $\ell$ satisfies
$$
\ell^2 = 1 + \sum_{k=1}^\infty \frac{2}{2^k} + \sum_{k=1}^\infty \frac{1}{4^ka_k^2} = 3 + \sum_{k=1}^\infty \frac{1}{4^ka_k^2} \tag{3}
$$
Since we have established that $1 \leq a_n \leq \ell$ for every $n\geq 1$, we get (also using in the LHS that $a_1=1$ to get better precision):
$$
3 + \frac{1}{4} + \frac{1}{\ell^2}\sum_{k=2}^\infty \frac{1}{4^k} \leq \ell^2 \leq 3 + \sum_{k=1}^\infty \frac{1}{4^k} \tag{4}
$$
that is, since $\sum_{k=1}^\infty \frac{1}{4^k}=\frac{1}{3}$,
$$
\frac{13}{12}+\frac{1}{12\ell^2} \leq \ell^2 \leq \frac{10}{3} \tag{5}
$$
which gives $\frac{39+\sqrt{1569}}{24}\leq \ell^2 \leq \frac{10}{3}$, i.e.,
$$
1.810\approx\boxed{\sqrt{\frac{39+\sqrt{1569}}{24}}\leq \ell \leq \sqrt{\frac{10}{3}}} \approx 1.828
$$
