<s>[INST] Is the following limit finite ....? I would like to see some clue for the following problem:
Let $a_1=1$ and $a_n=1+\frac{1}{a_1}+\cdots+\frac{1}{a_{n-1}}$, $n>1$. Find
$$
\lim_{n\to\infty}\left(a_n-\sqrt{2n}\right).
$$
 [/INST] Quick and dirty proof.
Assume for a moment that the sequence $\,a_n\,$ is a continuous
and differentiable function $\,a(n)\,$ of $\,n\,$ as is suggested in the picture below.

Then consider the following sequence. Fasten your seatbelts!
$$
a_{n+1} = a_n + \frac{1}{a_n} \\
\frac{a_{n+1} - a_n}{1} = \frac{1}{a_n} \\
\frac{a(n+dn) - a(n)}{dn} = \frac{1}{a(n)} \\
a'(n)\,a(n) = 1 \\ \frac{da^2(n)}{dn} = 2 \\
a^2(n) = 2n+C
$$
[Scaling argument deleted. I see no way to get it consistent, let it be rigorous. See edits of this post]
Now we would like to have $\,C=0$ ; and fortunately there is a boundary condition, for $\,n=2$ , that fits the bill : $\,C = a_2^2 - 2\cdot 2 = 0$ . So:
$\;a_n = \sqrt{2n}$ .
The larger $n$ , the better all of the approximations. A much neater way to express the end-result is:
$$
\large \boxed{\lim_{n\to\infty} \frac{a_n}{\sqrt{2n}} = 1}
$$
BONUS. Lemma:
$$
a_{n+1} = a_n + \frac{1}{a_n} \quad \Longrightarrow \quad
a_{n-1}^2 - a_na_{n-1} + 1 = 0 \quad \Longrightarrow \quad a_{n-1} = \frac{a_n}{2} + \sqrt{\left(\frac{a_n}{2}\right)^2-1}
$$
Now compute (the discretization of) the second derivative, assuming again that $\,a_n\,$ is large:
$$
a''(n) = a_{n+1} - 2a_n + a_{n-1} = a_n + \frac{1}{a_n} - 2a_n + \frac{a_n}{2} +
\frac{a_n}{2}\sqrt{1-\frac{1}{(a_n/2)^2}} \approx \\ \frac{1}{a_n} - \frac{a_n}{2} + \frac{a_n}{2}
\left[1-\frac{1}{2}\frac{1}{(a_n/2)^2} - \frac{1}{8}\left(\frac{1}{(a_n/2)^2}\right)^2\right] = -\frac{1}{a_n^3}
$$
Which means that the discrete function $\,a_n\,$ for large $\,n\,$ is actually very smooth , when seen as a continuous
and differentiable $\,a(n)$ . This is an even stronger motivation for the above treatment.It is noticed that the "true" derivatives exhibit the same structure as the discretizations:
$$
\left(\sqrt{2x}\right)'' = \left(\frac{1}{\sqrt{2x}}\right)' = -\frac{1}{\left(\sqrt{2x}\right)^3}
$$
There are no coincidences in mathematics.
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