Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$ Let $a_1=2$ and we define $a_{n+1}=a_n+\sqrt {a_n},n\geq 1$.
Is it possible to get a good aproximation of the $n$th term $a_n$?
The first terms are $2,2+\sqrt{2}$, $2+\sqrt{2}+\sqrt{2+\sqrt{2}}$ ...
Thanks in advance!

For the third term, we have the following. Let $c_n = \sqrt{a_n} -\frac{1}{2}n + \frac{1}{4}\ln n$, then $c_{n+1} -c_{n} = -\frac{1}{2} \cdot \frac{\sqrt{a_n}}{(\sqrt{a_n}+\sqrt{\sqrt{a_n}+a_n})^2} + \frac{1}{4} \ln(1+ \frac{1}{n})\quad (1)$. Note that $a_n = \frac{1}{4}n^2 - \frac{1}{4}n\ln n + o(n \ln n)$, the first term in R.H.S of (1) equals to $-\frac{1}{8} \cdot \frac{2n - \ln n + o(\ln n)}{n^2 -n \ln n + o(n \ln n)} = -\frac{1}{8n} \cdot(2+ \frac{ \ln n + o(\ln n)}{n - \ln n + o( \ln n)} )= -\frac{1}{4n} - \frac{\ln n}{8n^2}(1+o(1))$. Therefore (1) $= - \frac{\ln n}{8n^2}(1+o(1))$. Then we have $\lim_{n \rightarrow \infty} \frac{c_{n+1} -c_{n} }{- \frac{\ln n}{8n^2}} = 1$ and also have $\lim_{n \rightarrow \infty} \frac{c_{n+1} -c_{n} }{ \frac{\ln (n+1)+1}{8(n+1)} -\frac{\ln n+1}{8n} } = 1$. For $\frac{\ln n+1}{8n}$, we may integrate $-\frac{\ln x}{8x^2}$ with respect to $x$.
