Let $x_{n+1} = x_n + 1/(x_1 + x_2 +\ldots + x_n)$ with $x_1 = 1$. Show that $x_n\sim\sqrt{2\log(n)}$.

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 $$ \sum_{k=2}^n\left(\sum_{j=1}^k\sqrt{2\log(j)}\,\mathcal{I}(j)\right)^{-1} =\sqrt{2\log(n)}\,\mathcal{I}(n)\tag{5} $$

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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 $$ x_{n+1}=\sqrt{2\log(n)}+O\left(\frac1{\sqrt{\log(n)}}\right)\tag{7} $$