Let $y_n=x_n^2-n$, we have $y_1=0,y_{n+1}=y_n+\frac{1}{4(n+y_n)}$. Therefore $0<y_n\le \frac 14 H_{n-1}=(1+\frac 12 +...+\frac{1}{n-1})/4<\ln(n)/4$. It give $\sqrt n <x_n<\sqrt{n+\ln n }<\sqrt n (1+\frac{\ln n}{n}).$ Exactly $y_n>\frac{1}{4(n-1+\ln(n-1)/4}+...+\frac{4}{4}>H_{n-1}/4-\frac{1}{16} \sum_{k=1}^{n-1} \frac{\ln(n)}{n^2}>\frac{\ln n -1}{4}.$ Therefore $n+\frac{\ln n -1}{8}<x_n\sqrt n <n+\frac{\ln n }{8}$. (ln(625)<8)

Very nice Rust .

We can show that $625<{x_{625}}^2<625+\frac{1}{4}+\frac{623}{9}$ by basic method.

With this sequence we can find out the general of sequence and use this to prove result. $ v_n=2x_n$ And consider the function: $ f(x)=\frac{x+\sqrt{2}}{x-\sqrt{2}}$

We wish to show that $25 \le x_{625} < 25 + \frac{1}{25}$. Consider the sequence $(y_n)_{n \ge 1}$, where $y_n = x_n^2$. Our recurrence relation then becomes \[ y_{n+1} = y_n + 1 + \frac{1}{4y_n}.\]Since $y_1 = 1$, it follows that \[ y_n = n + \sum_{i = 1}^{n-1} \frac{1}{4y_i}. \]We see immediately that $y_n \ge n$ for all $n$. Thus $625 \le y_{625}$, so $25 \le x_{625}$. Now we need to show that $x_{625} < 25 + \frac{1}{25}$. This is equivalent to $y_{625} < 625 + 2 + \frac{1}{625}$. In fact, we will show that $y_{625} \le 625 + 2$; the desired conclusion follows. Note that $y_{625} = 625 + \sum_{i=1}^{624} \frac{1}{4y_i}$. Since $y_i \ge i$, we see that $\frac{1}{4y_i} \le \frac{1}{4i}$, so \[ y_{625} \le 625 + \sum_{i=1}^{624} \frac{1}{4i}. \]It suffices to show that $\sum_{i=1}^{624} \frac{1}{4i} \le 2$, or equivalently \[ \sum_{i=1}^{624} \frac{1}{i} \le 8. \] Let $H_n$ denote $\sum_{i=1}^n \frac{1}{i}$, so we want to show that $H_{624} \le 8$. We have the following Lemma: Lemma. $\log (n+1) < H_n \le 1 + \log n$ Proof. Note that $H_n = \sum_{i=1}^n \frac{1}{i} > \int_1^{n+1} \frac{1}{t}\, dt = \log (n+1)$ and $H_n = 1 + \sum_{i=2}^n \frac{1}{i} \le 1 + \int_1^n \frac{1}{t} \, dt = 1 + \log n$. $\blacksquare$ Now using Lemma, we see that $H_{624} \le 1 + \log 624$, so it suffices to show that $\log 624 \le 7$. Thus it suffices to show that $\log 625 \le 7$. Since $\log 625 = 4 \log 5$, we wish to show that $\log 5 \le \frac{7}{4}$, or equivalently $e^{7/4} \ge 5$. Now we use the Taylor expansion of $e^x$; namely, \[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots \]This means that \[ e^{7/4} > 1 + (\frac{7}{4}) + \frac{(7/4)^2}{2} + \frac{(7/4)^3}{6} + \frac{(7/4)^4}{24}. \]We easily evaluate the RHS as \[1 + (1 + \frac{3}{4}) + (1 + \frac{17}{32}) + \frac{343}{384} + \frac{2401}{6144}\]\[ = 3 + (\frac{3}{4} + \frac{2401}{6144}) + (\frac{17}{32} + \frac{343}{384})\]\[ > 3 + 1 + 1 = 5,\]as desired. $\blacksquare$