I'm trying to follow the following calculation of the definite integral using a limit of Riemann Sums. I have a question on the reasoning for a step in derivation of the limit which I will detail.
The original post is here.
$$\int_0^3 x^3 dx = \lim \limits_{n \to \infty} \sum_{i=1}^n f(x^*_i) \Delta x$$
$$= \lim \limits_{n \to \infty} \sum_{i=1}^n f\left(\frac{3i}{n}\right)\left(\frac 3n\right)$$
$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right) \sum_{i=1}^n \left(\frac{3i}{n}\right)^3$$
$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right)^4 \sum_{i=1}^n i^3$$
My first question is how is $\left(\frac{n(n+1)}{2}\right)^2$ multiplied to both sides when the RHS is unknown. Is the RHS assumed to be 1?
$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right)^4 \left(\frac{n(n+1)}{2}\right)^2 \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2$$
My second question is how are the next two equations derived from the prior?
$$= \left(\frac {3^4} 4\right) \lim \limits_{n \to \infty} \left(\frac 1 {n^4}\right) \left(\frac{n(n+1)}{2}\right)^2$$
$$= \left(\frac {3^4} 4\right) \lim \limits_{n \to \infty} \left(1 + \frac 1 n\right)^2$$
$$= \frac {3^4} 4$$
Best Answer
$= \lim \limits_{n \to \infty} (\frac 3 n)^4 (\frac{n(n+1)}{2})^2 \sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$ should be
$= \lim \limits_{n \to \infty} (\frac 3 n)^4 (\frac{n(n+1)}{2})^2$ makes use of $\sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$.
(Also, I saw Riemann spelled incorrectly there.)
Addendum to answer additional question posted as comment on this question:
It should be
$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right)^4 \left(\frac{n(n+1)}{2}\right)^2 = \left(\frac {3^4} 4\right) \lim \limits_{n \to \infty} \left(\frac 1 {n^4}\right) \left(\frac{n(n+1)}{\color{red}1}\right)^2,$$
because $$\left(\dfrac3n\right)^4 \left(\dfrac{n(n+1)}{2}\right)^2=3^4\left(\dfrac1n\right)^4\dfrac1{2^2}(n(n+1))^2 .$$
Finally,
$$\left(\dfrac1n\right)^4(n(n+1))^2=\dfrac{n^2}{n^4}(n+1)^2=\dfrac1{n^2}(n+1)^2=\left(\dfrac{n+1}n\right)^2=\left(1+\dfrac1n\right)^2.$$