MHB Why Does the Summation Use n-1 and i² in This Limit Calculation?

[karush]

11.2 nmh{2000}
Find a formula for the sum of $n$ terms
Use the formula to find the limit as $n\to\infty$

$\displaystyle \lim_{n\to\infty}

\sum\limits_{i = 1}^{n}\frac{1}{n^3}(i-1)^2=

\displaystyle \lim_{x\to\infty}\frac{1}{n^3}

\sum\limits_{n = 1}^{n-1}i^2$

This was from an solution to the problem but i didn't understand the $n-1$ on top of the $\Sigma$ or how they got the $i^2$ from the given. there are more steps but ?? about this one
thnx ahead

my try at LateX today lots of previewing

 

[QuestForInsight]

They have replaced i with i+1. It's sort of like a substitution for sums.

Suppose we let k = i-1, then k = 0 when i = 1 and n-1 when i = n, thus we have:

$\displaystyle \lim_{n \to \infty}\frac{1}{n^3}\sum_{i=1}^{n}(i-1)^2 = \lim_{n \to \infty}\frac{1}{n^3}\sum_{k=0}^{n-1}k^2 $

But k is a dummy variable so we can replace it with, say, i, and we have:

$\displaystyle \lim_{n \to \infty}\frac{1}{n^3}\sum_{k=0}^{n-1}k^2 = \lim_{n \to \infty}\frac{1}{n^3}\sum_{i=0}^{n-1}i^2 = \lim_{n \to \infty}\frac{1}{n^3}\sum_{i=1}^{n-1}i^2$

Where in the last step we wrote the sum from i = 1 because the term at i = 0 is 0.

 

[karush]

well that makes sense... they just didn't mention anything about it...

 

[QuestForInsight]

By the way, this limit is the Riemann sum of x² over [0, 1]:

$\displaystyle \lim_{n \to \infty}\frac{1}{n^3}\sum_{i=1}^{n-1}i^2 = \lim_{n \to \infty}\frac{1}{n^3}\sum_{i=1}^{n}i^2 = \int_{0}^{1}x^2\;{dx} = \frac{1}{3}.$

 

[karush]

https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy