<s>[INST] 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 $\displaystyle \lim_{n \rightarrow \infty} (x_n-n)=\frac{1}{4}$
 [/INST] My answer is wrong, but I would like to know why.
$$n(x_n-n-\frac{1}{4})=n \sum (\sqrt{(1+\frac{k}{n^2})}-(1+\frac{1}{4n}))$$
$$=n \sum \frac{(1+\frac{k}{n^2})-(1+\frac{1}{4n} )^2}{
\sqrt{(1+\frac{k}{n^2})}+1+\frac{1}{4n}
}$$
$$=n \sum \frac{
\frac{k}{n^2} - \frac{1}{16n^2} -\frac{1}{2n}}{
\sqrt{(1+\frac{k}{n^2})}+1+\frac{1}{4n}
}
$$
Since $$
\sqrt{(1+\frac{1}{n^2})}+1+\frac{1}{4n} \leq \sqrt{(1+\frac{k}{n^2})}+1+\frac{1}{4n}  \leq \sqrt{(1+\frac{n}{n^2})}+1+\frac{1}{4n}
$$, 
and they have the same limit as $n$ tends to infinity, we will apply sandwich theorem to the sum later.
Now, 
$$n \sum (\frac{k}{n^2} - \frac{1}{16n^2} -\frac{1}{2n} )$$
$$=n (\frac{n(n+1)}{2n^2} - \frac{1}{16n} -\frac{1}{2} )$$
$$=(\frac{n+1}{2} - \frac{1}{16} -\frac{n}{2} )$$
$$=(\frac{1}{2} - \frac{1}{16})=7/16$$
So even after dividing two, because of the denominator, it is $7/32$...
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