Summation (Part Two, Method Two)

To read Method One click here.

What is $1 - 2 + 3 - 4 + 5 - ...$ equal?

Let's call the sum $s$ .

$2s = (1 - 2 + 3 - 4 + ...) + (1 - 2 + 3 - 4 + ...)$

$2s = 1 + (- 2 + 3 - 4 + 5 - ...) + 1 - 2 + (3 - 4 + 5 - 6 + ...)$

$2s = 0 + (- 2 + 3) + (3 - 4) + (- 4 + 5) +...$

$2s = 1 - 1 + 1 - 1 + 1 ...$

$2s = 1 + (- 1 + 1) + (- 1 + 1) + ...$

$2s = 1 + 0 + 0 + 0 + ...$

$s = \frac {1}{2}$

But, from my previous method, the answer was $\frac {1}{4}$ !

How is this possible?

#Summation #NumberFallacy

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Note that $1-1+1-1+\cdots=\dfrac{1}{2}$ , via another famous summation. Therefore $2s=\dfrac{1}{2}$ , and $s=\dfrac{1}{4}$ as desired. $\Box$

Daniel Liu - 7 years, 2 months ago

This is because of Rienmann series thereom. This sequence is conditionally convergent so rienmann series theorem states that it could be rearranged in order to get any number. http://en.wikipedia.org/wiki/Riemannseriestheorem

Ashar Tafhim - 7 years, 5 months ago

What you have done here is mathematically incorrect. You can't pair up terms like that. As a matter of fact on the line $2s=1-1+1-1+1+...+(-1)^n$ is a geometric series with $r=-1$ and the sum diverges, i.e. it does not sum up to a specific value (A geometric series only converges when $|r|<1$ ). The correct method to solve this problem is to pair up the terms as:

$s = 1-2+3-4+5...=\lim_{n \rightarrow \infty}(1+3+5+...+(2n+1))-(2+4+6+...+(2n))=$

$\lim_{n \rightarrow \infty}(n^2)-(n^2+n)=\lim_{n \rightarrow \infty}-n=-\infty$ .

As you can see the sum diverges (tends to $-\infty$ ) and so is undefined.

Muhammad Shariq - 7 years, 5 months ago

I'd like to add that the way your summing the series, it is even possible to get 1/8/, 1/16....and so on as the answer which is clearly absurd.

Muhammad Shariq - 7 years, 5 months ago

However, this series yields a finite sum, which is clearly defined, because of a unique property of infinity.

Nanayaranaraknas Vahdam - 7 years ago

yes it is possible . you should search rienmann series theorem which states that if a series is conditionally convergent it can always be re arranged to get any number.

Ashar Tafhim - 6 years ago

The series of $1-1+1-1+1\ldots$ is the infamous Grandi series, which yields the sums $0, 1$ and $\frac12$ . However, for practical purposes, $\frac12$ is the accepted value, as $1$ and $0$ can be yielded by just changing the parentheses.

Nanayaranaraknas Vahdam - 7 years ago

Again, using the same method in your previous link that I mentioned, the partial sums diverge, so the sum also diverges. The partial sums for the first 2k terms is -1, for the first 2k+1 terms is (k+1)/2. Clearly the partial sums diverge.

Yong See Foo - 7 years, 5 months ago

Hey Yong See, you remember me from camp?

Sharky Kesa - 6 years, 1 month ago

There are two possibilities:1)the one shown in the note.2)the,series of( 1s )(-1s) ending with a -1 then the answer would be 0.

Adarsh Kumar - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$