find the limit of a sequnce I need to find the limit:
$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {{{(a + {1 \over n})}^2} + {{(a + {2 \over n})}^2} + ... + {{(a + {{n - 1} \over n})}^2}} \right]$  
any ideas here? I've tried to use "squeeze theorem" but with no luck..  
 A: If we add an $a^2$ to the bracket, we get a lower Riemann (or Darboux) sum for the integral
$$\int_a^{a+1} x^2\,dx,$$
and if we add an $(a+1)^2$, we get an upper sum, so
$$\int_a^{a+1} x^2\,dx - \frac{(a+1)^2}{n} \leqslant \frac1n \left[\sum_{k=1}^{n-1} \left(a+\frac{k}{n}\right)^2\right] \leqslant \int_a^{a+1} x^2\,dx - \frac{a^2}{n}.$$
The limit is therefore
$$\int_a^{a+1} x^2\,dx.$$
A: Let $u_n={1 \over n}\left[ {{{(a + {1 \over n})}^2} + {{(a + {2 \over n})}^2} + ... + {{(a + {{n - 1} \over n})}^2}} \right]$.
To elaborate on Daniel Fischer’answer : the Riemann sum approach yields
$u_n=\sum_{k=1}^{n-1} \frac{1}{n}f(\frac{k}{n})$ where $f(x)=(a+x)^2$, so $(u_n)$ converges to $\int_0^1 f(x)dx=\frac{(a+1)^3-a^3}{3}=\frac{3a^2+3a+1}{3}$.
Also, the expansion approach yields
$$
\begin{array}{lcl}
u_n &=& \frac{1}{n}. \sum_{k=1}^{n-1} (a+\frac{k}{n})^2 \\
  &=& \frac{1}{n}. \sum_{k=1}^{n-1} \big(a^2+2\frac{ak}{n}+\frac{k^2}{n^2}\big) \\
  &=& (\frac{n-1}{n})a^2+\frac{2a}{n^2}\big(\sum_{k=1}^{n-1}k\big)
  +\frac{1}{n^3}\big(\sum_{k=1}^{n-1}k^2\big) \\
  &=& (\frac{n-1}{n})a^2+\frac{a(n-1)}{n}
  +\frac{1}{n^3}\big(\frac{n(n-1)(2n-1)}{6}\big)
\end{array}
$$
And we see again that $(u_n)$ converges to $\frac{3a^2+3a+1}{3}$.
