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I am trying to solve this problem:

$$\lim_{n \to \infty}\sum_{j=1}^n {(a - {2j\over n})^3\over n}$$

My first instinct was to do a substition $x = {1 \over n}$ and rewrite the problem as: $$\lim_{x \to 0}\sum_{j=1}^\infty {(a - {2jx})^3}x$$ Here it looks obvious that this is a Riemann sum that can be evaluated as an integral, but I'm stuck. I don't know how to calculate this as an integral. This is the best I could come up with: $$\int_1^\infty {(a - {2jx})^3}dx$$ But I don't think it makes sense to evaluate at infinity for the upper bound. I think my integral is written wrong, but I'm not sure how to fix it. Any hints as to how to approach this problem properly would be appreciated.

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3 Answers 3

up vote 5 down vote accepted

Hint: a "right" Riemann sum for $\int_0^1 f(x)\ dx$ would be $$\sum_{j=1}^n \frac{f(j/n)}{n}$$ Your sum looks like that; what is the function $f$?

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Hint:

When $j$ changes from $1$ to $n$, then the term $\dfrac{j}{n}$ varies from $\dfrac{1}{n}$ to $1$, which leads to $0\leqslant x \leqslant 1.$

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\begin{align} \left(a - \dfrac{2j}n \right)^3 & = a^3 - 3a^2\left( \dfrac{2j}n\right) + 3a\left( \dfrac{2j}n\right)^2 - \left( \dfrac{2j}n\right)^3\\ \sum_{j=1}^n \left(a - \dfrac{2j}n \right)^3 & = na^3 - 3a^2\left( \dfrac{n(n+1)}n\right) + 3a\left( \dfrac{2n(n+1)(2n+1)}{3n^2}\right) - \left( \dfrac{2n^2(n+1)^2}{n^3}\right)\\ & = na^3 - 3a^2(n+1) + a\left( \dfrac{2(n+1)(2n+1)}{n}\right) - \left( \dfrac{2(n+1)^2}{n}\right)\\ \dfrac{\displaystyle \sum_{j=1}^n \left(a - \dfrac{2j}n \right)^3}n & = a^3 - 3a^2(1+1/n) + a\left( 2(1+1/n)(2+1/n)\right) - \left( 2(1+1/n)^2\right) \end{align} Hence, your limit is $$a^3 - 3a^2 + 4a-2 = (a-1)(a^2-2a+2)$$

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+1 - I find this solution very elegant because it doesn't use integration. –  hesson Dec 13 '12 at 8:51

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