Riemann Summation

Question

I am having a trouble with evaluating following problem.

$\lim_{n \to \infty} \frac{3}{n} \sum_{i=1}^n \left[\left(\frac{3i}{n}\right)^2-\left(\frac{3i}{n}\right)\right]$

Answer 1

Given $$\displaystyle \lim_{n\rightarrow \infty}\frac{3}{n}\sum^{n}_{i=1}\left[9\cdot \left(\frac{i}{n}\right)^2-3\cdot \left(\frac{i}{n}\right)\right] = \lim_{n\rightarrow \infty}\frac{9}{n}\sum^{n}_{i=1}\left[3\cdot \left(\frac{i}{n}\right)^2- \left(\frac{i}{n}\right)\right]$$

Now Using Reinmann Sum of Integrals, Put $\displaystyle \frac{i}{n} = x\;,$ Then $\displaystyle \frac{1}{n} = dx$ and

for upper and lower limit , $\displaystyle x_{l} = \lim_{n\rightarrow \infty}\frac{1}{n} = 0$ and $\displaystyle x_{u} = \lim_{n\rightarrow \infty}\frac{n}{n} = 1$

So our Sum convert into $$\displaystyle I = 3\int_{0}^{1}(3x^2-x)dx = \frac{3}{2}$$

Answer 2

Try rewriting the summand to get

$$\frac9{n^2}(1^1 + 2^2 + ... +n^2) - \frac3n(1 + 2 +3 +...+ n)$$ Then use the usual formulas for these sums to collapse it to a formula. Multiply by the other factor (3/n) outside the sigma sign. You should get a rational expression with a quadratic in both numerator and denominator. Taking the limit of this as n goes to infinity should give an answer.