Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$

Question

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!

Answer 1

Here is a method to develop the asymptotic behaviour of $a_n$ with a precision of two terms.

First, simply remark that $a_n$ grows to infinity since $a_n \geq n$ (by induction).

Then remark that $\sqrt{a_{n+1}} - \sqrt{a_n} = \frac{a_{n+1}-a_{n}}{\sqrt{a_{n+1}} + \sqrt{a_n}} = \frac{1}{1+\sqrt{1+\sqrt{\frac{1}{a_n}}}} \rightarrow \frac{1}{2}$ since $a_n\rightarrow\infty$.

It follows that $\sqrt{a_n} - 2=\sum_{i=0}^{n-1}\sqrt{a_{i+1}} - \sqrt{a_i}=\frac{1}{2}n+\mathcal{o}(n)$, then $a_n=(\frac{n}{2}+\mathcal{o}(n))^2 = \frac{n^2}{4}+\mathcal{o}(n^2)$

For the next term, we plug the asymptotic development we have just obtained into $\frac{1}{1+\sqrt{1+\sqrt{\frac{1}{a_n}}}}$ and look for an equivalent of $\frac{1}{1+\sqrt{1+\frac{1}{\sqrt{\frac{n^2}{4}+\mathcal{o}(n^2)}}}}-\frac{1}{2} \sim \frac{1-\sqrt{1+\frac{1}{\sqrt{\frac{n^2}{4}+\mathcal{o}(n^2)}}}}{4} \sim \frac{1}{4}\frac{1}{2}\frac{1}{\sqrt{\frac{n^2}{4}+\mathcal{o}(n^2)}} \sim \frac{1}{4n}$

From this, we deduce $\sqrt{a_{n+1}} - \sqrt{a_n} = \frac{1}{2}-\frac{1}{4n}+\mathcal{o}(\frac{1}{n})$, then $\sqrt{a_n} = \frac{n}{2}-\frac{1}{4}S_n + \mathcal{o}(S_n)$ where $S_n=\sum_0^n\frac{1}{k}=\ln(n)+\mathcal{o}(\ln(n))$.

It follows that $a_n=(\frac{n}{2}-\frac{1}{4}\ln(n)+\mathcal{o}(\ln(n)))^2=\frac{n^2}{4}-\frac{n\ln(n)}{4}+ \mathcal{o}(n\ln(n))$.

I believe that plugging again this new asymptotic development into the first formula would produce subsequent terms.

Answer 2

Yes, the approximation is $$a_n\sim \frac{n^2}{4}+o\left(n^2\right).$$ An heuristic derivation that should be possible to make rigorous is just replace $a_n$ by a smooth function $a(n)$. Then the recurrence relation gives a differential equation for $a(n)$: $$a_{n+1}-a_n\to a'(n)=\sqrt{a(n)}\quad \Longrightarrow \quad a(n)=\frac{(n+C)^2}{4}.$$

Answer 3

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 $x$.