That's rational!

If we let $S = \sqrt {n + \sqrt {n + \sqrt {n + \cdots}}}$ , we also get that $S = \sqrt {n + S}$ . Squaring both sides, rearranging to find $n$ and factorising, we get $n = S(S - 1)$ . To find the largest value of $n$ , we want the largest value of $S$ . Since $\sqrt{4000000} = 2000$ , the values of $S$ and $S - 1$ can't be $2001$ and $2000$ , but the largest values are $2000 \times 1999 = \boxed{3998000}$

Let, $y=\sqrt { n+\sqrt { n+\sqrt { n+............ } } }$

$\Rightarrow y=\sqrt { n+y } \\ \Rightarrow { y }^{ 2 }=n+y\\ \Rightarrow { y }^{ 2 }-y=n\\ \Rightarrow y(y-1)=n$

Here, $n$ is multiplication of $2$ consecutive number. Now we take the highest limit for $n$ and root it we find,

$\sqrt { 4000000 } =2000$

Now we take $1999$ and $2000$ to multiply to find the largest $n$

And the largest $n=1999\times 2000=\boxed { 3998000 }$

Let $s = \sqrt {n + \sqrt {n + \sqrt {n + \cdots}}}$ , then it's true that $s = \sqrt{n + s}$ . The solution of a quadratic equation for $s$ is $\frac{1 \pm \sqrt{1+4n}}{2}$ , and because $s$ is supposed to be rational, then it must also be true that $1 + 4n = a^2$ (where $a$ is a rational number). By simplifying the expression we get $n = \frac{a^2 - 1}{4}$ . For $n$ to be whole, $a$ also has to be whole, because if $a = \frac{p}{q}$ is a fraction in it's basic form, then $n = \frac{p^2 - q^2}{4q^2}$ is supposed to be a whole number, and so $q$ would need to divide $p$ , which is in controversy with $\frac{p}{q}$ in it's basic form. By further simplifying we get $n = \frac{a-1}{2} \times \frac{a+1}{2}$ which shows that $a$ has to be odd, and for $n \le 4000000$ , $\frac{a+1}{2} = 2000$ . From which $n = 1999 \times 2000 = \boxed{3998000}$