Is there a general method to find the asymptotic order for this sequence? Given $$a_{n+1}=a_n+\frac{n}{a_1+\dots+a_n},\qquad a_1>0$$ The answer is $$\lim_{n\to\infty} a_n\sim\sqrt{3}\cdot\sqrt{n}-\frac{\sqrt{3}}{4}\cdot\frac{1}{\sqrt{n}}$$
It is easy to show this sequence is increasing, and is divergent, because if assume the opposite $A=\lim_{n\to\infty} a_n$, we will get $A=A+\frac{1}A$, which gives contradictions. We also have
$$a_{n+1}\ge a_n +\frac{1}{a_n}\ge2$$
From here I can show:
$$a_1+\frac{n}{a_n}\le a_{n+1}\le a_1+\frac{n}{a_1}$$
Update.(1)
This equation can be also written as:
$$\begin{align}
a_n&=\sum_{k=1}^n a_k-\sum_{k=1}^{n-1} a_k\\
\\
a_n&=\frac{n}{a_{n+1}-a_n}-\frac{n-1}{a_{n}-a_{n-1}}\\
\\
a_n(a_{n+1}-a_n)(a_{n}-a_{n-1})&=n(a_{n}-a_{n-1})-(n-1)(a_{n+1}-a_n)\tag{1}
\end{align}$$
If assume $a_n=c\cdot n^p$
$$a_{n+1}-a_n\sim cp\cdot n^{p-1},~~~a_{n}-a_{n-1}\sim cp\cdot n^{p-1}$$
Plug into $(1)$ and only keep the leading order:
$$c^2p\cdot n^{2p-1}=1$$
So we get $p=\frac{1}2$ and $c=\sqrt{2}$
Why the leading order coefficient is $\sqrt{2}$, not $\sqrt{3}$?
The bottom line is, if I take the answer as template, let $$a_n=c\sqrt{n}+t\frac{1}{\sqrt{n}}+o(\frac{1}{\sqrt{n}})$$
Pretend we don't know coefficients $c$ and $t$. Now we want to compute $c$ and $t$. To the leading order approximation, We have
$$a_{n+1}-a_n= c\cdot \frac{1}{2\sqrt{n}}+O(\frac{1}{n^{3/2}})$$
Next, plug into Eq.$(1)$ and we can solve for $c=\sqrt{2}$. Why does this give a contradiction?
Update.(2)
Thank you to @Sangchul Lee , @Somos and @Youem
I put the computation part in the answer box below, and it works for asymptotic approximation at arbitrary order.


$$a_{n+1}=a_n+\frac{n}{\sum_{k=1}^n a_k}$$

This equation can be written as:
$$\sum_{k=1}^n a_k=\frac{n}{a_{n+1}-a_n}$$
Further, we can find $a_n$ by taking subtraction:
$$\begin{align}
a_n&=\sum_{k=1}^n a_k-\sum_{k=1}^{n-1} a_k\\
\\
a_n&=\frac{n}{a_{n+1}-a_n}-\frac{n-1}{a_{n}-a_{n-1}}\\
\\
a_n(a_{n+1}-a_n)(a_{n}-a_{n-1})&=n(a_{n}-a_{n-1})-(n-1)(a_{n+1}-a_n)\tag{1}\\
\\
&=(a_{n+1}-a_n)-n(a_{n+1}-2a_n+a_{n-1})
\end{align}$$
First determine the leading order. Let $a_n=c_0\cdot n^p$
$$\begin{align}
a_{n+1}&=c_0(n+1)^p=c_0 \left(n^p+pn^{p-1}+\frac{1}{2}p(p-1)n^{p-2}+...\right)\tag{2}\\
\\
a_{n-1}&=c_0(n-1)^p=c_0 \left(n^p-pn^{p-1}+\frac{1}{2}p(p-1)n^{p-2}+...\right)\tag{3}
\end{align}$$
Plug into $(1)$ and only keep the leading order on both sides:
$$c_0n^p\cdot c_0pn^{p-1}\cdot c_0pn^{p-1}=(2 c_0 p - c_0 p^2)n^{p-1}$$
So we get:
$$3p-2=p-1,~~~~~~c_0^3 p^2=(2 c_0 p - c_0 p^2)$$
$$\Rightarrow p=\frac{1}2,~~~c_0=\sqrt{3}$$
Next, find the next order, so we set the solution Ansatz as:
$$a_n=\sqrt{n}\left(c_0+c_1\cdot\frac{1}n \right),~~~~~c_0=\sqrt{3}$$
Plug into $(1)$ and re-do the series expansion to update $(2)$ and $(3)$. Now, comparing the coefficients for $\frac{1}{n^{3/2}}$
$$-\frac{3}4c_1\frac{1}{n^{3/2}}=\left(-\frac{\sqrt{3}}8 - \frac{5}4c_1\right)\frac{1}{n^{3/2}}$$
$$\Rightarrow c_1=-\frac{\sqrt{3}}{4}$$
Next, set the solution Ansatz as:
$$a_n=\sqrt{n}\left(c_0+c_1\cdot\frac{1}n+c_2\cdot\frac{1}{n^2} \right),~~~~~c_0=\sqrt{3},~~c_1=-\frac{\sqrt{3}}{4}$$
Now, comparing the coefficients for $\frac{1}{n^{5/2}}$
$$\frac{3}{32}(\sqrt{3}-40c_2)\frac{1}{n^{5/2}}=\frac{3}{64}\left(\sqrt{3} - 112c_2\right)\frac{1}{n^{5/2}}$$
$$\Rightarrow c_2=-\frac{\sqrt{3}}{32}$$
Keep going on this method, we get:
$$a_n=\sqrt{n}\left(\sqrt{3}-\frac{\sqrt{3}}{4}\cdot\frac{1}n-\frac{\sqrt{3}}{32}\cdot\frac{1}{n^2}-\frac{\sqrt{3}}{128}\cdot\frac{1}{n^3}-\frac{79\sqrt{3}}{38912}\cdot\frac{1}{n^4}+\cdots \right)$$
