# Trouble understanding Big O notation for a sum of n integers [duplicate]

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This problem is an example in a Discrete Math textbook. How can big-O notation be used to estimate the sum of the first n positive integers?

Solution: Because each of the integers in the sum of the first n positive integers does not exceed n, it follows that $$1+2+...+n \le n+n+...+n=n^2$$ From this inequality it follows that $1+2+...+n$ is $O(n^2)$.

I do not understand how the sum of $n$'s is equal to $n^2$. Can anyone explain to me what the book used to get this?

## marked as duplicate by Antonio Vargas, ml0105, M Turgeon, apnorton, user61527 May 1 '14 at 17:16

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• I'm confused as to why you'd want to estimate the sum of the first n positive integers since this is a well-known formula $\frac{n(n+1)}{2}$. – EgoKilla May 1 '14 at 1:54
• How many "$n$"s are in the sum "$n+n+\cdots+n$"? – Eric Towers May 1 '14 at 1:55

## 2 Answers

The sum of $n$ $n$'s is $n \cdot n$, by definition. $n \cdot n = n^2$.

Also, Big-O notation is used for bounding, not estimation. As Big-O notation removes all constant multipliers and merely represents an upper asymptotic bound, it is not a useful estimator.

• Maybe "\cdot" ? – Eric Towers May 1 '14 at 1:56
• Done. I am always too lazy to make that switch in my homework, but it's probably a bad habit I should break out of. – user2258552 May 1 '14 at 1:57
• Thank you! This was an example problem in Rosen's Discrete Math book. I didn't really understand the purpose of the example but just needed to clarify that one step. – Kot May 1 '14 at 1:59
• @user2258552 "Big-O notation removes all constant multipliers" was the key to understand this, thanks! – rgkobashi Jan 27 at 20:30

First, you note that$$1+2+\ldots+n = \frac{n^2+n}{2}$$ Since this is a quadratic polynomial, follows that it is $O(n^2)$.