I'm confused by this problem. Is the answer 1 or 0 OR 1/2? If the

[saratchandra]

I'm confused by this problem. Is the answer 1 or 0 OR 1/2? If the series is upto n, then the answer is 1 if the number is odd and 0 if even. If you assume that the answer is x, then 1-x = x and the answer is 1/2. Is the answer 1 or 0 OR 1/2? or simply indeterminate?

 

[dextercioby]

In mathematics such a series is called conditionally convergent. It means that it matters in which order you are summing its terms. You can write your series in 3 different ways that eventually lead to 3 different answers, 0,1,1/2.

 

[disregardthat]

No, the value of this series does not depend on the order of the terms, but on the terms themselves. What is each term? Is it 1,-1,1,-1,... ? In that case it diverges. Is it 1-1,1-1,1-1,... i.e. 0,0,0,0,..? In that case it converges. Any other definition of the terms?

The order of the terms 1,-1,1,-1,... doesn't matter in an infinite series, it will diverge anyway in that case.

 

[Edgardo]

Hi saratchandra, welcome to PF!

You have already recognized that you have a series. Can you write down the series with a summation sign? And how can you check whether a series is convergent or divergent?

 

[D H]

If you assume that the answer is x, then 1-x = x and the answer is 1/2.

To demonstrate what saratchandra is talking about, let S be the value of this series,

$$S=\sum_{r=0}^{\infty} (-1)^r$$
Then if S exists, we can compute 1-S by adding 1 and the additive inverse of S:

$$1-S=1-\sum_{r=0}^{\infty} (-1)^r = 1+\sum_{r=0}^{\infty} (-1)(-1)^r<br /> = 1+\sum_{r=1}^{\infty}(-1)^r = \sum_{r=0}^{\infty} (-1)^r = S$$
or 1-S=S, yielding S=1/2. This is of course illegal; just because an operation is valid when applied to a convergent series does not mean it can be applied to a divergent series. However, this value of 1/2 appears in many other forms when evaluating this series. For example,

$$\frac 1{1+x} = \sum_{r=0}^{\infty} (-1)^rx^r$$
Evaluating this at x=1 yields

$$\frac 1{1+1} = 1/2 = \sum_{r=0}^{\infty} (-1)^r$$
Hah! 1-1+1-1+... = 1/2! Yet another place this series arises is in the eta function,

$$\eta(z) = \sum_{r=1}^{\infty} \frac{(-1)^{r+1}}{r^z}$$
At z=0, this becomes

$$\eta(z) = 1/2 = \sum_{r=0}^{\infty} (-1)^r$$
So once again, 1-1+1-1+... = 1/2. Another way to look at the eta function is via its relation to the zeta function,

$$\eta(z) = (1-2^{1-z})\zeta(z)$$
Thus $\eta(0) = -\zeta(0)$. The series representation of the zeta function is

$$\zeta(z) = \sum \sum_{r=1}^{\infty} \frac{1}{r^z}$$
Evaluating this series at z=0 yields

$$\zeta(0) = -1/2 = 1+1+1+1+\cdots$$
This shows how the series 1-1+1-1+... is related to the series 1+1+1+1+..., which every physicist knows has a value -1/2 (zeta function renormalization).

Some very nasty tricks were employed here, several of them illegal. On the other hand, analytic continuity is a very legal operation, and this was used underneath the hood to find several of these evaluations. Several divergent series can indeed take on well-defined, consistent values when analyzed in the context of analytic continuity.

 

[saratchandra]

Thanks a lot DH for the detailed explanation. Thanks to others too

 

[GenePeer]

This is of course illegal; just because an operation is valid when applied to a convergent series does not mean it can be applied to a divergent series. However, this value of 1/2 appears in many other forms when evaluating this series. For example,

$$\frac 1{1+x} = \sum_{r=0}^{\infty} (-1)^rx^r$$
Evaluating this at x=1 yields

$$\frac 1{1+1} = 1/2 = \sum_{r=0}^{\infty} (-1)^r$$
Hah! 1-1+1-1+... = 1/2!

That is illegal yet as well. The geometric progression works only if 0<x<1. As it appears that there is no legal way to prove this, would it not be better to say the sum is undefined?

 

[disregardthat]

That is illegal yet as well. The geometric progression works only if 0<x<1. As it appears that there is no legal way to prove this, would it not be better to say the sum is undefined?

What do you mean by better? The sum does $$\sum^{\infty}_{n=0} x^{n}$$ not exist as a limit of partial sums for x = -1, but one can extend the domain of definition as one please. In this particular extension (which is the unique analytical extension to C-{1} I believe), we will have that it equals 1/2.

 

[GenePeer]

I meant "safer". Anyway, so what you mean is we define the sum to be 1/2 because it "extends" properly from previous know formulae? Like how we define 0! = 1, even though the factorial definition is for positive integers only?

 

[disregardthat]

I meant "safer". Anyway, so what you mean is we define the sum to be 1/2 because it "extends" properly from previous know formulae? Like how we define 0! = 1, even though the factorial definition is for positive integers only?

Yes, exactly like how we define 0! = 1, even though the recursive definition of n! as n! = n*(n-1)! and 1! = 1 only applies to n>0.

In this case however the extension is less "arbitrary", in the sense that there is only one analytical extension of the function 1/(1-x) (on the open ball of radius 1 in C) to C-{1}. In other words it is only one particular way of extending the function to C-{1} while preserving differentiability (as a complex function).

 

[logics]

No, the value of this series does not depend on the order of the terms, but on the terms themselves.

I am not an expert, pardon me if I venture a layman opinion: the value of the series depends on the number of terms:
if it is even = 0, if it's odd = 1
infinity is even or odd?

 

[D H]

By almost all the techniques taught in lower level math classes (sophomore college and below) to test for whether a series is convergent, series such as 1-1+1-1+... , 1+1+1+1+..., 1+2+3+4+..., etc. fail to converge. All of them fail the simplest test of all: the terms that comprise the series do not converge to zero.

So you can stop right there and say these series are divergent, they have no value -- or you can go to more advanced techniques. There are a variety of ways to give meaning to divergent series. Cesàro summation, Abel summation, Borel summation, ... You do have to give some things up along the way. For example, order doesn't matter in an absolutely convergent series. Order does matter with these divergent series. Take Grandi's series, 1-1+1-1+... Swap adjacent even (2n) and odd(2n+1) terms and you get -1+1-1+1... Per those more advanced techniques, this is a different series, one whose value is -1/2.

Yet another technique is that of analytic continuation. Suppose some function f(z) has a series representation f(z)=Ʃa_nϕ_n(z) for z within some domain (e.g., a radius of convergence r about some point z₀). Suppose the function f(z) has an analytic continuation outside of that domain. Because the analytic continuation of a function (if it exists) is unique, it makes a lot of sense to say that the value of the series at some point outside the radius of convergence of the series is the value of the analytic continuation of the series at that point.

 

[Dickfore]

If you accept that 1 - 1 + 1 - 1 + ... = 1/2, then you also have to accept that

1 - 2 + 3 - 4 + ... = 1/4

Proof:
Look at the series:
$$S(x) = \sum_{n = 1}^{\infty}{x^{n}}$$
and find its derivative:
$$S'(x) = \sum_{n = 1}^{\infty}{n x^{n - 1}}$$
So, our series in question is $S'(-1)$. But, if you look at $S(x)$, it is actually a geometric series with a sum:
$$S(x) = \frac{1}{1 - x}$$
Although the sum itself is strictly convergent for $|x| < 1$, this rational expression is defined for all $x \neq 1$ (We say that the rational expression is an analytic continuation of the function defined in the unit circle by the convergent series). Then, take the derivative:
$$S'(x) = \frac{1}{(1 - x)^2}$$
and you get $S'(-1) = 1/4$. This method is called Abel summation.

Notice that the above formulas fail for $x = 1$, i.e. you cannot say anything about, for example, the series

1 + 1 + ...

and

1 + 2 + ...

 

[disregardthat]

I am not an expert, pardon me if I venture a layman opinion: the value of the series depends on the number of terms:
if it is even = 0, if it's odd = 1
infinity is even or odd?

The value of an infinite series is the limit of the sequence of partial sums (which are both even and odd). As you say, they alternate between 0 and 1, and so will not converge at all.

Infinity is not an integer, it's not even nor odd.

 

[logics]

. As you say, they alternate between 0 and 1, and so will not converge at all. Infinity is not an integer, it's not even nor odd.

I was joking, disregard that, that's what I meant: not convergent, not divergent, alternating: 0/1 , value could be either 0 or 1. Beyond technicalities the bottom line is: value is not decidable

 

[TGlad]

Is there anything wrong with saying that the sum of the sequence 1,-1,1,-1,... converges to the two values 0 and 1? i.e. a multi-valued or double peak convergence.

If we put our foot down and just say 'undefined' or 'undecidable' then we seem to be missing out on informing people about the limit. For instance, what does e^ix converge to as x tends to +infinity? Would it not be better to say that it converges to the set of points on the unit circle than to just say it is undecidable.

 

[disregardthat]

Is there anything wrong with saying that the sum of the sequence 1,-1,1,-1,... converges to the two values 0 and 1? i.e. a multi-valued or double peak convergence.

If we put our foot down and just say 'undefined' or 'undecidable' then we seem to be missing out on informing people about the limit. For instance, what does e^ix converge to as x tends to +infinity? Would it not be better to say that it converges to the set of points on the unit circle than to just say it is undecidable.

Well, you can say that 1 and 0 are the only values for which a subsequence of the partial sums can converge to. As you mention as well, the suplim and inflim of the sequence of partial sums converge to 1 and 0 respectively.

But there is no mis-information about saying that the series diverges. In fact it is instructive to say so, because I think many people might think of diverging series exclusively as unbounded series, or even series that grows indefinitly.

 

[logics]

As you say, they alternate between 0 and 1, and so will not converge at all..

As I said, I am considering the problem from a logical point of view: if argument is valid to prove it does not converge, should be valid to prove it does not diverge. A swindle to prove this has not yet been presented, but it is possible. The concept of convergence/divergence implies "change" of value, direction: here values do not change at all, start 0/1 and stay 0/1 at infinity. Definition, rules should allow exceptions for cases like Grandi: no convergence , no divergence, no Cesaro sum. Just a layman opinion.

 

[disregardthat]

As I said, I am considering the problem from a logical point of view: if argument is valid to prove it does not converge, should be valid to prove it does not diverge. A swindle to prove this has not yet been presented, but it is possible. The concept of convergence/divergence implies "change" of value, direction: here values do not change at all, start 0/1 and stay 0/1 at infinity. Definition, rules should allow exceptions for cases like Grandi: no convergence , no divergence, no Cesaro sum. Just a layman opinion.

I couldn't make any sense of what you are talking about here. The fact of the matter is that the series does not converge.

 

[D H]

As I said, I am considering the problem from a logical point of view: if argument is valid to prove it does not converge, should be valid to prove it does not diverge. A swindle to prove this has not yet been presented, but it is possible. The concept of convergence/divergence implies "change" of value, direction: here values do not change at all, start 0/1 and stay 0/1 at infinity. Definition, rules should allow exceptions for cases like Grandi: no convergence , no divergence, no Cesaro sum. Just a layman opinion.

What swindle?

Convergence and divergence are flip sides of the same freshman calculus coin. Analyzing a series from the perspective of the convergence/divergence of the sequence of partial sums is but one way to try to assign a value to that series. The best way to interpret "divergence" is that these elementary techniques are not powerful enough assign a value to the series in question. Other techniques can assign a value. Any two techniques that are regular and linear that do assign some value to some series will agree with one another.

This is just one of many cases where the simple techniques taught in freshman/sophomore mathematics are not up to some task. Another example is integration. Consider the indicator function, I_Q(x) = 1 if x is rational, 0 if x is irrational. What is the integral of this function from x=0 to x=1?

 

[logics]

What swindle??

Eilenberg-Mazur

 

[256bits]

This may be elementary but why can't it be solved this way:

S=∑r=0∞(−1)r

S = 1 - 1 + 1 - 1 + 1 - 1 + . . .

S = 0 + 1 - 1 + 1 - 1 + 1 - 1 + . . .

Add
2S = 1

S = 1/2

 

[disregardthat]

This may be elementary but why can't it be solved this way:



S = 1 - 1 + 1 - 1 + 1 - 1 + . . .

S = 0 + 1 - 1 + 1 - 1 + 1 - 1 + . . .

Add
2S = 1

S = 1/2

It doesn't work because there is not such S. In other words, it's nonsensical (in the sense of adding an infinite series with another).

 

[256bits]

It doesn't work because there is not such S. In other words, it's nonsensical (in the sense of adding an infinite series with another).

Reason I asked is that is how some of the proofs in high school math were formulated.
Can't remeber if they were divergent or convergent series.

 

[disregardthat]

Reason I asked is that is how some of the proofs in high school math were formulated.
Can't remeber if they were divergent or convergent series.

It is perfectly fine if the series converge. It rests on the fact that if a_n --> a, b_n --> b, then a_n+b_n --> a+b.

 

[D H]

This post contains replies to several recent posts.

Eilenberg-Mazur

Huh? Except for the connection made in very poorly written (and apparently biased) wikipedia article on the Eilenberg-Mazur swindle, there is no connection between the Grandi series and this swindle. The article equates fallacious proofs of 1=0 via the Grandi series to the Eilenberg-Mazur swindle. This is fallacious reasoning in and of itself. For one thing, the Eilenberg-Mazur swindle is valid.

For another, there's nothing special about the Grandi series here. Consider the alternating harmonic series, 1-1/2+1/3-1/4+... This is a convergent series: It converges to ln(2). Now let's rearrange the terms of the series. The algorithm for the rearrangement is a state machine that alternates between two states:

• In state 1 (the initial state), find the first remaining additive term (e.g., 1, 1/3, 1/5, ...). Pair this with the subtractive term that is half of the first term. Add the sum of the two as a new term in the rearranged series. Eliminate both terms in the original formulation from further consideration.
• In state 2, find the first remaining subtractive term. Add this term as a new term in the rearranged series and eliminate the found term in the original formation from further consideration.

The rearranged series is (1-1/2)-1/4+(1/3-1/6)-1/8+(1/5-1/10)-1/12+(1/7-1/14)-1/16+... = 1/2-1/4+1/6-1/8+1/10-1/12+1/14-1/16 = 1/2*(1-1/2+1/3-1/4+...). By rearranging the terms in the series we have halved the value of the series! In fact, given any number, it is possible to find a rearrangement of the alternating harmonic series that yields that number as a sum.

The problem here is that alternating harmonic series is a conditionally convergent series; it is not absolutely convergent. This magic of rearranging terms to come up with drastically different sums applies to any conditionally convergent series. What this means is that rearranging terms is invalid for such series. (It is a valid operation for absolutely convergent series.)

This may be elementary but why can't it be solved this way:

S = 1 - 1 + 1 - 1 + 1 - 1 + . . .

S = 0 + 1 - 1 + 1 - 1 + 1 - 1 + . . .

Add
2S = 1

S = 1/2

The problem here is that you are manipulating terms of an alternating series that is not absolutely convergent. It is an illegal operation.

It doesn't work because there is not such S. In other words, it's nonsensical (in the sense of adding an infinite series with another).

There certainly is such an S. You just can't find it using elementary techniques. Cesaro summation, Abel summation, Borel summation, and zeta function regularization all yield the same answer, 1/2, to 1-1+1-1+...

It is perfectly fine if the series converge. It rests on the fact that if a_n --> a, b_n --> b, then a_n+b_n --> a+b.

As noted above, this is not a valid operation on conditionally convergent series.

 

[disregardthat]

There certainly is such an S. You just can't find it using elementary techniques. Cesaro summation, Abel summation, Borel summation, and zeta function regularization all yield the same answer, 1/2, to 1-1+1-1+...

Of course, I meant that $$\lim_{N \to \infty} \sum^N_{n=0} (-1)^n$$ does not exist. It should be obvious from the context.

As noted above, this is not a valid operation on conditionally convergent series.

It most certainly is valid. Note that I don't allow a rearrangement of the terms as you propose in your post. It's a matter of adding finite partial sums with well defined terms, something I earlier have stressed. Now I'm talking about converging series, not zeta function regularization or something else.

 

[logics]

... For one thing, the Eilenberg-Mazur swindle is valid...

that's right, that is what I meant in my post #18:
scientists may debate for centuries ("liar" paradox [true/false]) when they are reluctant to accept the principle that "exception proves the rule"

...For another, there's nothing special about the Grandi series here...

Grandi has some analogy [convergent/divergent] with Russell's http://wikipedia.org./wiki/Present_King_of_France" [true/false], if you consider it as the antecedent of Leibniz [a= -2]
$$\sum^{\infty}_{n=0} [a=-1]^{n}$$
It ought to be re-written, re-defined.
Is 1,0,0,0,... a series? is it convergent? is it divergent? what is Cesaro-sum?

the sense of post #18 was:
[so expresses logical implication $\Rightarrow$] : disregardthat has made a valid argument:

... they alternate between 0 and 1, and so will not converge at all...

values alternate $\Rightarrow$ series is not convergent. [$\Rightarrow$...]
This is a new argument, because swindle [is valid] had already proven that, but had the consequence ("tertium non datur") that series is convergent.

This is a better argument, as it applies to both "possibilities" and we must not wait for another swindle. If it is so, disregardthat has made a swindle-preventing discovery.

 

[pwsnafu]

$$\sum^{\infty}_{n=0} [a=-1]^{n}$$

You need to define your notation. I've never seen that.

Is 1,0,0,0,... a series? is it convergent? is it divergent? what is Cesaro-sum?

It is convergent sequence, and not a series.
Cesaro is http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation" .