Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,030

Grandi's series is the infinite sum: 1-1+1-1...

What's interesting about it is that it cna have several 'values':

1. We can group the terms like this:

(1-1)+(1-1)...=0+0+0+0+0+...=0

2. We can group the terms this way:

1+(-1+1)+(-1+1)...=1+0+0+...=1

3. The final way that I know of is to have s be equal to the sum:

s=1-1+1-1...

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

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

s=1-s

2s=1

s=1/2

which is quite an unusual answer.

http://en.wikipedia.org/wiki/Grandi's_series

*Last edited by anonimnystefy (2011-06-17 05:28:48)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

Offline

**soroban****Member**- Registered: 2007-03-09
- Posts: 452

. . .

. . .

. . .

. . .

. . .

. .

. . . .**

. .

.

Offline

**MathPro101****Member**- Registered: 2011-06-08
- Posts: 18

Question: When I explained Grandi's series to a friend, they stated that addition is commutative, which is true. They wanted to know why the commutative property is allowed to be violated for this series. I suggested that it has to do with the fact that it is a series, and not just any normal addition. Could anyone help me explain why we don't break any laws? Thank you.

Offline

**JohnA****Member**- Registered: 2016-01-13
- Posts: 3

To me this looks like an easier explanation. Commutative law isn't broken in any way (as far as I can see).

```
S = 1 - 1 + 1 - 1 + 1 ...
-1 + S = -1 + 1 - 1 + 1 - 1 ...
adding:
S + -1 + S = 0 + 0 + 0 + 0 + 0 ... = 0
-1 + 2S = 0
2S = 1
S = 1/2
```

No worries about having different infinite lengths (sequence "-1 + S" is exactly one longer than sequence S) and so the two sequences line up nicely all the way to infinity. No need to invoke different ways to group the terms at all. And no need to say it has "several values".

Offline

**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 243

and

Offline

**JohnA****Member**- Registered: 2016-01-13
- Posts: 3

Stangerzv wrote:

and

Interesting! But there's something a little odd in there ...

Here's my reasoning (not sure if it's true):

- For a finite series, the length of the series (aka cardinality) of "1 to 2n" is different than "1 to 2n+1" for any n you choose.

- But for an infinite series, the length of "1 to 2n" and "1 to 2n+1", as n approaches infinity, is the same (according to Georg Cantor)

- The two infinite lengths are the same since you can draw a 1 - 1 correspondence from each term in the sequence "1 to 2n" to each term in the sequence "1 to 2n + 1" - as long as n goes to infinity! Stopping anywhere along the way breaks the 1 - 1 correspondence, so you can't stop.

- In your two formulas the value you arrive at (0 or 1) is depending on the sequences having different lengths, but in fact both sequences have the same infinite length.

- Another way to look at it, is that n has hit infinity before the "=" sign is reached

I think the correct formulas are:

and

and you can add whatever you want to the summation sign, the limit to infinity overrides it:

John

Offline

**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 243

You get funny values when you take all the infinities as the same. Infinity as a concept gives rise to many problems like S=1-1+1-...infinity=1/2 because we tend to take all the infinities as the same like (1+infinity)=infinity. What is the value of infinity/infinity? I am more to the concept that the Infinity has an origin and point of reference. For example, if we had a source of light that could travel forever into the infinity and is set to travel today and the another one would be set to travel tomorrow, are they the same value when they reach infinity?

If they do have the same value, I can show you that Riemman zeta function

not and many more.*Last edited by Stangerzv (2016-01-16 12:07:29)*

Offline

Grandi’s series is divergent. With divergent series, and also conditionally convergent series, you get different “answers” by grouping the terms in different ways.

Another example with a divergent series:

One way of grouping the terms gives

Another way gives

No doubt you can get other results with other groupings.

*Last edited by Nehushtan (2016-01-17 05:13:20)*

Offline

**JohnA****Member**- Registered: 2016-01-13
- Posts: 3

Nehushtan wrote:

Grandi’s series is divergent. With divergent series, and also conditionally convergent series, you get different “answers” by grouping the terms in different ways.

I agree, Grandi's Series is divergent. And I agree that you get different answers by grouping in different ways.

But please notice I didn't "group the terms" in Grandi's Series as you say. What I did was I lined the terms up for two infinite series (one was Grandi's series, the other was a slightly modified version of the same series). The two series are the same length because they have a simple 1 - 1 correspondence which lasts all the way to infinity.

I think it is quite possible to get different answers like you stated:

- if you group terms or

- if you don't have a 1-1 correspondence (i.e. you skip a term, or clump a a couple of terms or more, or... something) or

- if you stop prematurely (i.e. you convert the infinite series to a finite series),

but I didn't do any of those things.

Could you show me a different answer for Grandi's Series other than 1/2? Of course, it would be best if you used the same technique I used i.e. lining up two infinite series using a demonstrable 1-1 correspondence. It would really help me to understand!

thanks,

John

PS I really don't know if my "proof" is right. I'm not insisting or anything like that.

It's just that I haven't broken any axioms or operational rules in Integer Algebra. As far as I know, each operation I'm using has no flaw. So, to me, the "proof" looks good.

PSS I do realize I am making these assumptions:

1) I assume that it is ok to sum a1 + a2 + a3... to infinity. As someone pointed out, "+" is defined as a binary operation, so it can only take two operands. So a1 + a2 + a3 ... is really a1 + (a2 + (a3 + .... ))))) <-- an infinite number of parentheses here -- which may cause a problem. I don't see it, but maybe there is one.

2) it's ok to sum the terms of two infinite series if they are the same length, and there's no shifting or misalignment issues way down the line, i.e. the two series start aligned and stay aligned to infinity.

3) the sum of 0+0+0... to infinity is 0. Again it seems reasonable, but see #1

4) algebraic manipulations like "2S = 1 --> divide both sides by 2 --> S = 1/2." are ok to do when S started off as an infinite series. I know these manipulations are ok to do if S is a simple, finite number. But it is possible, I guess, that if S actually has the value "infinity" then you do an operation like "2S=1 --> divide both sides by 2", it will result/report as a simple, finite number. It seems very, very unlikely that this would be the case, but I don't know for sure.

5) others?

If you guys see a flaw in any step or operation I've done or any other assumptions, please let me know!

Offline

Pages: **1**