# [SOLVED]-11.2 why n-1 on summation

#### karush

##### Well-known member
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

my try at LateX today lots of previewing

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#### QuestForInsight

##### Member
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.

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#### karush

##### Well-known member
well that makes sense... they just didn't mention anything about it...

#### QuestForInsight

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

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