Let $x_{n+1} = x_n + 1/(x_1 + x_2 +\ldots + x_n)$ with $x_1 = 1$. Show that $x_n\sim\sqrt{2\log(n)}$. As the title states we have a sequence defined by
$$x_{n+1} = x_n + \dfrac{1}{x_1 + x_2 + \cdots + x_n}$$
with $x_1 = 1$.
The first few terms are: $1, 2, \frac{7}{3}, \frac{121}{48} \cdots$
Any ideas would be appreciated.
 A: Repeatedly integrating by parts yields
$$
\int\sqrt{2\log(x)}\,\mathrm{d}x=x\sqrt{2\log(x)}-x\sum_{k=0}^{n-1}\frac{(2k-1)!!}{\sqrt{2\log(x)}^{\,2k+1}}-\int\frac{(2n-1)!!\,\mathrm{d}x}{\sqrt{2\log(x)}^{\,2n+1}}\tag{1}
$$
and
$$
\int\frac{\mathrm{d}x}{\sqrt{2\log(x)}}=\frac{x}{\sqrt{2\log(x)}}+x\sum_{k=1}^{n-1}\frac{(2k-1)!!}{\sqrt{2\log(x)}^{\,2k+1}}+\int\frac{(2n-1)!!\,\mathrm{d}x}{\sqrt{2\log(x)}^{\,2n+1}}\tag{2}
$$
Therefore, using the Euler-Maclaurin Sum Formula and $(1)$ and $(2)$, for $n\gt1$, we get
$$
\left(\sum_{j=1}^n\sqrt{2\log(j)}\,\mathcal{I}(j)\right)^{-1}=\frac1{n\sqrt{2\log(n)}}\,\mathcal{I}(n)\tag{3}
$$
where $\mathcal{I}(n)=1+O\left(\frac1{\log(n+1)}\right)$.
Furthermore,
$$
\int\frac{\mathrm{d}x}{x\sqrt{2\log(x)}^{\,k}}=\frac1{2-k}\sqrt{2\log(x)}^{\,2-k}+C\tag{4}
$$
Thus, applying the Euler-Maclaurin Sum Formula to $(3)$ and $(4)$ gives
$$
\sum_{k=2}^n\left(\sum_{j=1}^k\sqrt{2\log(j)}\,\mathcal{I}(j)\right)^{-1}
=\sqrt{2\log(n)}\,\mathcal{I}(n)\tag{5}
$$

Using the definition of $x_n$, we get
$$
\begin{align}
x_{n+1}
&=2+\sum_{k=2}^nx_{k+1}-x_k\\
&=2+\sum_{k=2}^n\left(\sum_{j=1}^kx_j\right)^{-1}\tag{6}
\end{align}
$$
Equation $(5)$ and $(6)$ show that
$$
x_{n+1}=\sqrt{2\log(n)}+O\left(\frac1{\sqrt{\log(n)}}\right)\tag{7}
$$
A: I'll rename the series $y_n$ for convenience. Think of $y_n$ as samples of the function $y(x)$, at points $x_n = n dx$.
Now, your relation simply states that:
$$y_{n+1}-y_{n}= y'(x)dx = \frac{dx}{\int y(x) dx}$$
So you have an integral equation that must be solved. I don't know how to solve it, but showing that your function satisfies the equation is easy:
$$\frac{d\sqrt{2\log x}}{dx}=\frac{1}{x\sqrt{2\log x}}$$
And for large enough x:
$$\int\sqrt{2\log x}\ dx\approx x\sqrt{2\log x}$$
A: I revisited this question and came up with a very different approach. It is so different, that I didn't think it could be incorporated into my previous answer; so, I have added a second answer.

Basic Bounds
Since $x_1=1$ and
$$
x_{n+1}-x_n=\left(\sum_{k=1}^nx_k\right)^{-1}\tag1
$$
$x_n$ is a positive, monotonically increasing, concave function of $n$. Therefore,
$$
\frac{n+1}2\,x_n\le\sum_{k=1}^nx_k\le nx_n\tag2
$$
The left-hand inequality is because $x_k\ge\frac knx_n$ ($x_1=1,x_2=2$, and $x_n$ is concave in $n$; concavity is maintained by extending $x_0=0$). The right-hand inequality is because $x_k\le x_n$.
We can modify the left hand inequality in $(2)$ by substituting $n\mapsto n+1$ and then subtracting $x_{n+1}$ from both sides to get
$$
\frac{n}2\,x_{n+1}\le\sum_{k=1}^nx_k\le nx_n\tag3
$$
Taking reciprocals of $(3)$ and applying $(1)$, we get
$$
\frac1{nx_n}\le x_{n+1}-x_n\le\frac2{nx_{n+1}}\tag4
$$
Noting that $2x_n\le x_n+x_{n+1}\le2x_{n+1}$, we can multiply $(4)$ by $x_n+x_{n+1}$ to get
$$
\frac2n\le x_{n+1}^2-x_n^2\le\frac4n\tag5
$$
Summing $(5)$ yields
$$
\sqrt{1+2H_n}\le x_{n+1}\le\sqrt{1+4H_n}\tag6
$$

Asymptotics
Apply $(4)$ and $(6)$ to get that $\frac{x_m}{x_n}\ge1+\epsilon$ requires $m-n\gt\frac{x_n\epsilon}{\frac2{nx_n}}\ge nH_n\epsilon$. Thus, if we set $m=n+nH_n\epsilon$, for $n\le k\le m$, we have $x_k\ge\frac{x_m}{1+\epsilon}$. Therefore,
$$
\begin{align}
\sum_{k=1}^mx_k
&\ge\frac{x_m}{1+\epsilon}\overbrace{\frac{mH_n\epsilon}{1+H_n\epsilon}}^{\substack{\text{number of}\\\text{terms}\\\ge\frac{x_m}{1+\epsilon}}}\tag7\\
&=mx_m\underbrace{\frac1{1+\epsilon}\frac{H_n\epsilon}{1+H_n\epsilon}}_{\substack{\text{can be made close to}\\\text{$1$ by choosing $\epsilon$ small}\\\text{and $n$ big}}}\tag8
\end{align}
$$
Due to $(8)$, for large $n$, $(3)$ becomes
$$
\sum_{k=1}^nx_k\sim nx_n\tag9
$$
which leads to $(6)$ becoming
$$
\begin{align}
x_{n+1}
&\sim\sqrt{1+2H_n}\tag{10}\\[3pt]
&\sim\sqrt{2\log(n)}\tag{11}
\end{align}
$$
