$\lim_{n\to \infty} \ \frac{1}{n} \Bigl[(a+\frac{1}{n})^2+(a+\frac{2}{n})^2+\cdots+(a+\frac{n-1}{n})^2\Bigr]$ without L'Hopital Find limit: 
$$a_n = \lim_{n\to \infty} \ \frac{1}{n} \Bigl[\Bigl(a+\frac{1}{n}\Bigr)^2+\Bigl(a+\frac{2}{n}\Bigr)^2+\cdots+\Bigl(a+\frac{n-1}{n}\Bigr)^2\Bigr]$$ 
I tried limiting it with $$n\cdot\frac{1}{n}\Bigl(a+\frac{1}{n}\Bigr)^2\leqslant a_n \leqslant n \cdot \frac{1}{n}\Bigl(a+\frac{n-1}{n}\Bigr)^2$$
but that got me to the answer that left side limits to $a^2$ and right side to $(a+1)^2$ 
and so I didn't squeeze it the right way. Help, and keep it simple.
 A: If you expand the squares, then you get
$$\begin{split} \sum_{k=1}^{n-1} \Big(a+\frac{k}{n}\Big)^2&=\sum_{k=1}^{n-1} \Big(a^2+2a\frac{k}{n}+\frac{k^2}{n^2}\Big)
\\
&=(n-1)a^2+2a\cdot \frac{1}{n}\cdot\frac{n(n-1)}{2}+\frac{1}{n^2}\cdot\frac{n(n-1)(2n-1)}{6}.
\\
&=(n-1)a^2+a(n-1)+\frac{(n-1)(2n-1)}{6n},
\end{split}$$
and thus
$$\begin{split} 
\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} \Big(a+\frac{k}{n}\Big)^2 &=\lim_{n\to\infty} \Big((1-1/n)a^2+a(1-1/n)+\frac{(n-1)(2n-1)}{6n^2}\Big)
\\
&=a^2+a+\frac{1}{3}.
\end{split}$$
A: Converting this
to a Riemann sum,
$\begin{array}\\
\frac{1}{n}\sum_{k=1}^n (a+\frac{k}{n})^2
&\to \int_0^1 (a+x)^2dx\\
&= \int_a^{a+1} x^2dx\\
&=\dfrac{x^3}{3}|_a^{a+1}\\
&=\dfrac{a^3+3a^2+3a+1-a^3}{3}\\
&=\dfrac{3a^2+3a+1}{3}\\
&=a^2+a+\frac13\\
\end{array}
$
A: $\lim_{n\to \infty} \ \frac{1}{n} \Bigl[\Bigl(a+\frac{1}{n}\Bigr)^2+\Bigl(a+\frac{2}{n}\Bigr)^2+\cdots+\Bigl(a+\frac{n-1}{n}\Bigr)^2\Bigr] =\displaystyle\int_{a}^{a+1} x^2\, dx=\dfrac{x^3}{3}|_a^{a+1}=\dfrac{(a+1)^3-a^3}{3}=a^2+a+\dfrac{1}{3}$
