limit of sum defined sequence Let $x_n=\displaystyle \sum_{k=1}^n \sqrt{1+\frac{k}{n^2}}, n\ge1$. Prove that $\displaystyle \lim_{n \rightarrow \infty} n (x_n-n-\frac{1}{4})=\frac{5}{24}$.
What I've done:it's easy to show that $\displaystyle \lim_{n \rightarrow \infty} \frac{x_n}{n}=1$ and $\displaystyle \lim_{n \rightarrow \infty} (x_n-n)=\frac{1}{4}$

Using that $\frac{k}{n^2} \xrightarrow[n\to\infty]{} 0$ for all $1\leq k\leq n$, one can use the Taylor expansion of $\sqrt{1+x}$, $$\sqrt{1+x} = 1+\frac{x}{2} - \frac{x^2}{8} + o(x^2)$$ when $x\to 0$:
$$\begin{align}
x_n &= \sum_{k=1}^n \sqrt{1+\frac{k}{n^2}}
= \sum_{k=1}^n \left(1+\frac{k}{2n^2}-\frac{k^2}{8n^4}+o\left(\frac{k^2}{n^4}\right)\right)
\\
&= n+ \frac{1}{2n^2}\cdot\frac{n(n+1)}{2} - \frac{1}{8n^4}\cdot\frac{n(n+1)(2n+1)}{6} + o\left(\frac{1}{n}\right)\\
&= n+ \frac{1}{4} + \frac{1}{4n} - \frac{1}{8n^4}\cdot\frac{2n^3}{6} + o\left(\frac{1}{n}\right)\\
&= n+ \frac{1}{4} + \frac{6}{24n} - \frac{1}{24n} + o\left(\frac{1}{n}\right)\\
&= n+ \frac{1}{4} + \frac{5}{24n} + o\left(\frac{1}{n}\right)
\end{align}$$
giving you $n(x_n - n - \frac{1}{4}) = \frac{5}{24} + o(1)$.
Note: Now, while the above is a valid and legit method, one should stop and at least consider why the manipulation and summation of the $o(\cdot)$'s is indeed acceptable. (Specifically: while it's implicit, what each $o(\cdot)$'s is taken with regard to? $n\to\infty$, or $\frac{k}{n}\to0$ each for a different $k$? And why is it still OK?)
