For $k\in\mathbb{R}$, evaluate $\lim\limits_{n \to \infty} (\sqrt{n^2+nk+1}-\lfloor \sqrt{n^2+nk+1} \rfloor)$ 
Let $a_n =\sqrt{n^2+nk+1}$ where $k$ is a real number. Find $$\lim_{n \to \infty} (a_n-\left\lfloor a_n \right\rfloor).$$

Using the Binomial Theorem, we can write:
$$\sqrt{n^2+nk+1}=n+\frac{k}{2}+O\left(\frac{1}{n}\right),\\
n+\frac{k}{2}+O\left(\frac{1}{n}\right)-\left\lfloor n+\frac{k}{2}+O\left(\frac{1}{n}\right) \right\rfloor \\
=\frac{k}{2}+O\left(\frac{1}{n}\right)-\left\lfloor \frac{k}{2}+O\left(\frac{1}{n}\right) \right\rfloor.$$
I'm not sure if this is the right approach or not.
 A: As should be clear, we are considering the limit for integer $n$; otherwise (i.e., if we examined $\lim_{x\to\infty}$ of the analogous function for real $x$) the limit won't exist. Note also that since we are restricting to integer $n$, the result is essentially periodic in $k$ with period $2$; the cases $k=-2,2,0$ are trivial and prevent true periodicity. Thus for $K\in\mathbb{R}\setminus\{-2,2,0\}$, it suffices to compute the limit for $k\in (0,2]$ with $K-k\in \mathbb{Z}$, in which case the limit is just $k/2$. Note that for even integers $K$ the limit is $1$ and $1/2$ for $K$ odd.
By definition, we can rewrite our terms as $$\sqrt{n^2+kn+1}-\lfloor n^2+kn+1\rfloor=\left\{\sqrt{n^2+kn+1}\right\}$$
$$
=\left\{n\sqrt{1+k/n+1/n^2}\right\}
$$Since $n$ is an integer, we can sneakily insert it here:
$$
=\left\{n\sqrt{1+k/n+1/n^2}-n\right\}
$$Now use an asymptotic for large $n$:
$$
=\left\{n\left(1+\frac{k}{2 n}+\frac{4-k^2}{8 n^2}+O\left(n^{-3}\right)\right)-n\right\}
$$
$$
=\left\{\frac{k}{2 }+\frac{4-k^2}{8 n}+O\left(n^{-2}\right)\right\}
$$We can choose $n$ large enough such that each of the error terms are negligible with respect to the floor function. This allows us to take the limit, which is $\{k/2\}$ (up to the details mentioned in the preamble).
See here for a similar analysis of the case $k=1$.
