$$a_n = \frac{1}{n^2} \sum _{k=1}^n \left( \sqrt{\left(n+k\right)^2+n+k}\, \right)$$
$$\lim_{n\to \infty} a_n =\:?$$
I solved it like this and I would want to know it this makes sense and if not why.
$$\frac{1}{n}\sum _{k=1}^n \left[ \sqrt{\frac{\left( n^2+2kn+k^2+n+k \right)}{n^2}} \; \right]= \frac{1}{n}\sum _{k=1}^n \left[ \sqrt{1+2\frac{k}{n}+\left( \frac{k}{n} \right)^2+ \frac{1}{n}+\frac{k}{n}\cdot\frac{1}{n}} \; \right]$$
and because $n$ tends to infinity, $\dfrac{1}{n}$ and $\dfrac{k}{n} \times \dfrac{1}{n}$ disappears. But remains
$$a_n=\frac{1}{n}\sum _{k=1}^n \sqrt{\left( \frac{k}{n}+1 \right)^2}$$
And so I considered a function $f:[0, 1] \to \mathbb{R}$, $f(x)=x+1$ and then I integrated for $x$ from $0$ to $1$ and got $\frac{3}{2}$, which should be the limit of $a_n$.
$$(n+k)^2+(n+k)=(n+k)(n+k+1)$$
For $k=1,2,\ldots,n$
$$(n+k)^2< (n+k)(n+k+1) <(n+k+1)^2$$
Now $$\frac{1}{n^2} \sum_{k=1}^{n} (n+k) <a_{n} < \frac{1}{n^2} \sum_{k=1}^{n} (n+k+1)$$
$$ \frac{1}{n^2} \left[ n^2+\frac{n(n+1)}{2} \right] <a_{n}< \frac{1}{n^2} \left[ n^2+\frac{n(n+3)}{2} \right]$$
$$ \frac{3}{2}+\frac{1}{2n} <a_{n}< \frac{3}{2}+\frac{3}{2n}$$
Take the limits for the upper and lower bounds and by sandwich theorem, we get $$\lim_{n\to \infty} a_{n} = \frac{3}{2}$$
Further points to be noticed
If your sum is running from $k=0$ to $k=n-1$, then your sum is exactly between the left and right rectangular rules of $\displaystyle \int_{0}^{1} (x+1)\, dx$.