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[SOLVED] why n-1 on summation

karush

Well-known member
Jan 31, 2012
2,725
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
 
Last edited:
Jul 22, 2012
35
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.
 
Last edited:

karush

Well-known member
Jan 31, 2012
2,725
well that makes sense... they just didn't mention anything about it...
 
Jul 22, 2012
35
By the way, this limit is the Riemann sum of x² over [0, 1]:

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