The limit floored me

Note that $n<\sqrt{n^2+n+1} < n+1$ Hence, $\lfloor \sqrt{n^2+n+1} \rfloor=n$ Hence, the given limit is $\lim_{n\to \infty} \sqrt{n^2+n+1}-n$ $=\lim_{n \to \infty} \frac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}=\frac{1}{2}$

We have $n^2 + n + 1 = n^2 \left( 1 + \frac1n + \frac1{n^2}\right)$ , then for positive $n$ , $\sqrt{n^2+n+1} = n \left(1 + \left(\frac1n+\frac1{n^2}\right)\right)^{\frac12}$ .

For $n$ to be sufficiently large, $\frac1n+\frac1{n^2}$ is very small, so we can apply the binomial expansion $(1+x)^N \approx 1 + Nx + \frac{N(N-1)}2 x^2$ for small $x$ . In this case, $x = \frac1n+\frac1{n^2}, N = \frac12$ . Then,

$\begin{array} { r c l} \sqrt{n^2 + n + 1}& =& n \left(1 + \left(\frac1n+\frac1{n^2}\right)\right)^{\frac12} \\ & \approx & n \left[ 1 + \frac12 \left(\frac1n+\frac1{n^2}\right) + \frac{\frac12 \left(\frac12 -1\right)}{2} \frac12 \left(\frac1n+\frac1{n^2}\right)^2 \right ] \\ & =& n + \frac12 + O\left(\frac1n\right) \\ \end{array}$

Where $O(n)$ denote the Big O notation. We're only interested in the fractional part of $\sqrt{n^2 + n + 1}$ , which is simply $\frac12 + O\left(\frac1n\right)$ . Take its limit yields $\boxed{\frac12}$ as the answer.

EClearly the limit is somewhere between 0 and 1. If you graph it, you can practically guess that the answer is .5, but here's the algebraic proof.

Begin by working with the floor function half of the equation. It can be observed that for all positive integer values of $n$ , $\left\lfloor\sqrt{n^2+n+1}\right\rfloor=n$ . Thus our function we are evaluating is $\displaystyle \lim_{n\rightarrow \infty} \sqrt{n^2+n+1}-n$ .

Next, Consider the perfect square polynomials $n^2$ and $n^2+2n+1$ and "average" the two. $\dfrac{n^2+n^2+2n+1}{2}=n^2+n+\frac{1}{2}$ . The radical part of the function we are evaluating is the average of two perfect squares (we can ignore the difference in constants because as $n$ achieves large enough values, the constant is negligible). So taking the average of our functions we have

$\sqrt{n^2}-n=0$

$\sqrt{n^2+2n+1}-n=1$

$\frac{0+1}{2}=\boxed{0.5}$