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
$$
\lim_{n\to\infty}\left(a_n-\sqrt{2n}\right).
$$

As other answers indicate, this sequence obeys to the following recurrence relation :
$$a_1=1\quad\mathrm{and}\quad\forall n\in\mathbb{N},\,a_{n+1}=a_n+\frac{1}{a_n}$$
This proves that the sequence is increasing. Supposing its convergence to some finite limit $L>0$ would lead to $L=L+\dfrac{1}{L}$, a contradiction.
Hence the sequence diverges towards $+\infty$.
Now, for all $n\in\mathbb{N}$, we have :
$$a_{n+1}^2-a_n^2=\left(a_n+\frac{1}{a_n}\right)^2-a_n^2=2+\frac{1}{a_n^2}\longrightarrow 2$$
By Cesaro's lemma :
$$\frac{1}{n}\left(a_n^2-a_0^2\right)=\frac{1}{n}\sum_{k=0}^{n-1}\left(a_{k+1}^2-a_k^2\right)\longrightarrow 2$$
Thus $$a_n\sim\sqrt{2n}$$ Now, we will use ...
Lemma Given a sequence $(u_n)_{n\ge1}$ of positive real numbers such that $u_n\sim\dfrac{1}{n}$, we have :
$$\sum_{k=1}^nu_k\sim\ln(n)$$
(See below for a proof.)
Since $a_{n+1}^2-a_n^2-2\sim\dfrac{1}{2n}$ and by the previous lemma :
$$a_n^2-2n\sim\frac{\ln(n)}{2}$$
which can be written :
$$a_n=\sqrt{2n}\,\sqrt{1+\frac{\ln(n)}{4n}+o\left(\frac{\ln(n)}{n}\right)}$$
Using now the Taylor expansion $\sqrt{1+t}=1+\frac{t}{2}+o(t)$ as $t\to0$, we get finally :
$$\boxed{a_n=\sqrt{2n}\left(1+\frac{\ln(n)}{8n}+o\left(\frac{\ln(n)}{n}\right)\right)}$$
In particular, we see that $\lim_{n\to\infty}\left(a_n-\sqrt{2n}\right)=0$, but the result above is much more accurate.

Proof of the above lemma
Given $\epsilon>0$, there exists $N\in\mathbb{N}^\star$ such that :
$$k>N\implies\left|u_k-\frac{1}{k}\right|\le\epsilon$$
As soon as $n>N$, we have :
$$\left|\sum_{k=1}^nu_k-\sum_{k=1}^n\frac{1}{k}\right|\le\underbrace{\left|\sum_{k=1}^N\left(u_k-\frac{1}{k}\right)\right|}_{=A}+\sum_{k=N+1}^n\left|u_k-\frac{1}{k}\right|\le A+\epsilon\sum_{k=N+1}^n\frac{1}{k}$$
And a fortiori :
$$\left|\sum_{k=1}^nu_k-\sum_{k=1}^n\frac{1}{k}\right|\le A+\epsilon\sum_{k=1}^n\frac{1}{k}$$
Since $\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k}=+\infty$, there exists $N'\in\mathbb{N}^\star$ such that :
$$n>N'\implies\sum_{k=1}^n\frac{1}{k}>\frac{A}{\epsilon}$$
Finally :
$$n>\max\{N,N'\}\implies\left|\sum_{k=1}^nu_k-\sum_{k=1}^n\frac{1}{k}\right|\le2\epsilon\sum_{k=1}^n\frac{1}{k}$$
This proves that :
$$\sum_{k=1}^nu_k\sim\sum_{k=1}^n\frac{1}{k}$$
But we know that $\sum_{k=1}^n\frac{1}{k}\sim\ln(n)$ as $n\to\infty$, hence the conclusion (by transitivity of $\sim$).
