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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..

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  • $\begingroup$ You can consider it as a Riemann sum (for which integral?), or you can expand the squares and treat the three groups of terms separately. The Riemann sum approach yields the answer with less computation. $\endgroup$ – Daniel Fischer Nov 2 '13 at 11:34
  • $\begingroup$ You could try setting $m=\frac{1}{n}$ then considering the limit as $m\to 0$. $\endgroup$ – Shaun Nov 2 '13 at 11:35
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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 $\frac{3a^2+3a+1}{3}$.

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  • $\begingroup$ hmmm.. according the last equality, the limit is: ${a^2} + a + {1 \over 3}$. isn't it? $\endgroup$ – captain dragon Nov 2 '13 at 12:56
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    $\begingroup$ @captaindragon $a^2+a+\frac{1}{3}$ is the same thing as $\frac{3a^2+3a+1}{3}$, so hmmm me no hmmms. $\endgroup$ – Ewan Delanoy Nov 2 '13 at 12:57
  • $\begingroup$ agreed. can you explain this: ${{3a{x^2} + 3{a^2}x + {a^3}} \over 3}$ $\endgroup$ – captain dragon Nov 2 '13 at 13:11
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    $\begingroup$ @captaindragon : corrected (deleted the x's, thanks). $\endgroup$ – Ewan Delanoy Nov 2 '13 at 13:15
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    $\begingroup$ @captaindragon It is. $\endgroup$ – Ewan Delanoy Nov 2 '13 at 13:19
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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

$$\int_a^{a+1} x^2\,dx.$$

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