Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

How can $\lim\limits_{n\to\infty} \sum_{i=1}^{n^2} 1$ result into 1+3+5...? Why odd numbers?

share|cite|improve this question

marked as duplicate by leonbloy, user17762, Amzoti, Martin Argerami, Sasha May 20 '13 at 3:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

$\lim_{n \to \infty}$ doesn't "result" in $1 + 3 + 5 + \cdots$. The limit doesn't exist. – 6005 May 20 '13 at 1:54
I mean, probably you are just talking about the partial sums, but if that's the case I don't understand why you put a limit sign there. – 6005 May 20 '13 at 1:55
@Goos as in Peter's context:… – user8005 May 20 '13 at 1:57
Peter's context involves a limit of ratios of partial sums, not a ratio of limits of partial sums. – anon May 20 '13 at 1:59
@user8005 I think that 1+3+5.. expresion is understood to be a limit – Gastón Burrull May 20 '13 at 2:00

Hoo boy. You're in for a treat; this theorem was known to Pythagoras and his disciples.

Theorem: The sum of the first $n$ odd numbers is equal to $n^2$.

Proof: By picture.

Picture proof for theorem

The original image is part of a page containing two proofs of this identity: It is not mine.

share|cite|improve this answer
I like this proof, essentially is the telescopic sum argument $\sum (2k-1)=\sum [k^2-(k-1)^2]=n^2$ – Gastón Burrull May 20 '13 at 2:41
AoPS has another picture, not exactly this one: – Gyu Eun Lee May 20 '13 at 2:46
Yeah I know them, but this is my favorite. – Gastón Burrull May 20 '13 at 2:47

$$1+3+5+\ldots+(2n-1)=\sum_{k=1}^n (2k-1)=\sum_{k=1}^n [k^2-(k-1)^2]=n^2-0^2=n^2=\sum_{k=1}^{n^2} 1$$

share|cite|improve this answer
Does this definition have a name? Your response on Goos's the limit doesn't exist? – user8005 May 20 '13 at 1:54
@user8005 I don't know, its just a variant of this sum – Gastón Burrull May 20 '13 at 1:55
@GastónBurrull Per your comment, regularization of divergent sums is a very tangential topic; I feel that bringing it up without clarifying the meaning of the original situation (see my comment above) with limits could be derailing. – anon May 20 '13 at 2:02
@anon I can only answer about partial sums, I think what OP means has nothing to do with the phenomenon to quantify divergent sums as… I think I'm ignorant about this other topic – Gastón Burrull May 20 '13 at 2:10

Note that the definition of summation $ \sum_{k=1}^m a_k $ $$ \sum_{k=1}^m a_k=a_1 +a_2+ \ldots +a_m. $$ So, for eachpositive integer $n$, we have $$ \sum_{k=1}^{n^2} 1=\sum_{k=1}^{n^2} (1+0k)=1+ 1+\ldots +1(n^2 {\rm times})=n^2. $$ On the other hand, for the same integer $n$, we have $$ {\rm sum\,\, of \,\,n-positive\,\, odd \,\,numbers}(=1+3+5+ \ldots+ (2n-1))=n^2. $$ Thus, we can write $$ \sum_{k=1}^{n^2} 1=1=3+5+\ldots+(2n-1) \tag 1 $$ Letting $n\to \infty$ in (1) we get the asked expression.

share|cite|improve this answer
Wow, $1=1+0k$, I never saw before such a trick, probably would help me working with hard sums – Gastón Burrull May 20 '13 at 2:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.