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
$$
\lim_{n\to\infty}\left(a_n-\sqrt{2n}\right).
$$
 A: 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 $\sim$).
A: Alternative solution: let us assume that there exists a real number $S $ such that 
$$ \lim_{n\to\infty} \left(a_n-S\sqrt{2n} \right)=0$$
If this constant (and necessarily positive) value exists, we should be able to determine it. On the other hand, if such constant value does not exist (for example, if $S $ is not a constant but a function of $n $), we could expect some contradiction.
Under the assumption that a real $S $ exists, we have 
$$  \lim_{n\to\infty}   \frac {1}{a_n}=  \lim_{n\to\infty}    \frac {1}{S \sqrt {2n}}$$
Also, because by definition the recurrence of the OP can be written as $a_{n+1}=a_n+1/a_n \,\, \,\,$, we have
$$  \lim_{n\to\infty}   \frac {1 }{a_n}=\lim_{n\to\infty} \left( a_{n+1}-a_n \right) \\ =\lim_{n\to\infty} \left( S \sqrt {2(n+1)} -S  \sqrt {2n}\right) $$
Combining the two equations above we get
$$\lim_{n\to\infty}  \frac {1}{S \sqrt {2n}}= \lim_{n\to\infty} S \left( \sqrt {2(n+1)} -  \sqrt {2n} \right)$$
Solving for $S $ and taking into account that $S$ cannot be negative,  we obtain
$$S= \lim_{n\to\infty}  \frac { \sqrt {n+1}  +  \sqrt {n}}{2 \sqrt {n} } = 1$$
Thus, assuming existence of real $S $, we obtain $S=1\,\,$ with no contradiction. We conclude that
$$ \lim_{n\to\infty} \left(a_n-\sqrt{2n} \right)=0$$
It should be pointed out that this solution does not give a proof that such limit exists, but only shows that, assuming its existence, it must be $1$.
A: The recurrence can be re-written as $a_{n+1}=a_n+\dfrac1{a_n}$. Let $x_n=a_n^2$; then we have $x_1=1$ and
$$
x_{n+1} = a_{n+1}^2 = \left(a_n+\dfrac1{a_n}\right)^2 = 
a_n^2+2+\dfrac1{a_n^2} = x_n+2+\dfrac1{x_n}.
$$
From this we can conclude $x_{n+1}>x_n+2$, so
$$ x_n \ge 2n-1; $$
then $x_{n+1}=x_n+2+\dfrac1{x_n}<x_n+2+\dfrac1{2n-1}$, so
$$ 
x_n < 2n-1 + \left(1+\dfrac13+\dfrac15+\ldots+\dfrac1{2n-3}\right)
< 2n+\log n.
$$
Finally,
$$
\big|a_n-\sqrt{2n}\big| = \dfrac{|a_n^2-2n|}{a_n+\sqrt{2n}} <
\dfrac{|x_n-2n|}{\sqrt{2n}} < \dfrac{\log n}{\sqrt{2n}}.
$$
Therefore,
$\big|a_n-\sqrt{2n}\big|\to0$.
A: 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.
