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- Thread starter saratchandra
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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.

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disregardthat

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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.

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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?

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To demonstrate what saratchandra is talking about, let S be the value of this series,If you assume that the answer is x, then 1-x = x and the answer is 1/2.

[tex]S=\sum_{r=0}^{\infty} (-1)^r[/tex]

Then

[tex]1-S=1-\sum_{r=0}^{\infty} (-1)^r = 1+\sum_{r=0}^{\infty} (-1)(-1)^r

= 1+\sum_{r=1}^{\infty}(-1)^r = \sum_{r=0}^{\infty} (-1)^r = S[/tex]

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,

[tex]\frac 1{1+x} = \sum_{r=0}^{\infty} (-1)^rx^r[/tex]

Evaluating this at x=1 yields

[tex]\frac 1{1+1} = 1/2 = \sum_{r=0}^{\infty} (-1)^r[/tex]

Hah! 1-1+1-1+... = 1/2! Yet another place this series arises is in the eta function,

[tex]\eta(z) = \sum_{r=1}^{\infty} \frac{(-1)^{r+1}}{r^z}[/tex]

At z=0, this becomes

[tex]\eta(z) = 1/2 = \sum_{r=0}^{\infty} (-1)^r[/tex]

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,

[tex]\eta(z) = (1-2^{1-z})\zeta(z)[/tex]

Thus [itex]\eta(0) = -\zeta(0)[/itex]. The series representation of the zeta function is

[tex]\zeta(z) = \sum \sum_{r=1}^{\infty} \frac{1}{r^z}[/tex]

Evaluating this series at z=0 yields

[tex]\zeta(0) = -1/2 = 1+1+1+1+\cdots[/tex]

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.

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

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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?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,

[tex]\frac 1{1+x} = \sum_{r=0}^{\infty} (-1)^rx^r[/tex]

Evaluating this at x=1 yields

[tex]\frac 1{1+1} = 1/2 = \sum_{r=0}^{\infty} (-1)^r[/tex]

Hah! 1-1+1-1+... = 1/2!

- #8

disregardthat

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What do you mean by better? The sum does [tex]\sum^{\infty}_{n=0} x^{n}[/tex] 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.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?

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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?

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disregardthat

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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).

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I am not an expert, pardon me if I venture a layman opinion: the value of the series depends on the number of terms:No, the value of this series does not depend on the order of the terms, but on the terms themselves.

if it is even = 0, if it's odd = 1

infinity is even or odd?

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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

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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:

[tex]

S(x) = \sum_{n = 1}^{\infty}{x^{n}}

[/tex]

and find its derivative:

[tex]

S'(x) = \sum_{n = 1}^{\infty}{n x^{n - 1}}

[/tex]

So, our series in question is [itex]S'(-1)[/itex]. But, if you look at [itex]S(x)[/itex], it is actually a geometric series with a sum:

[tex]

S(x) = \frac{1}{1 - x}

[/tex]

Although the sum itself is strictly convergent for [itex]|x| < 1[/itex], this rational expression is defined for all [itex]x \neq 1[/itex] (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:

[tex]

S'(x) = \frac{1}{(1 - x)^2}

[/tex]

and you get [itex]S'(-1) = 1/4[/itex]. This method is called Abel summation.

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

1 + 1 + ...

and

1 + 2 + ...

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- #14

disregardthat

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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.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?

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

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I was joking, disregard. As you say, they alternate between 0 and 1, and so will not converge at all. Infinity is not an integer, it'snoteven nor odd.

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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

- #17

disregardthat

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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.

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.

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.

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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.As you say, they alternate between 0 and 1, andso will not convergeat all..

- #19

disregardthat

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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.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.

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What swindle?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.

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

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256bits

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

S = 1 - 1 + 1 - 1 + 1 - 1 + . . .S=∑r=0∞(−1)r

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

Add

2S = 1

S = 1/2

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disregardthat

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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).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

- #24

256bits

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Reason I asked is that is how some of the proofs in highschool math were formulated.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).

Can't remeber if they were divergent or convergent series.

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disregardthat

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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.Reason I asked is that is how some of the proofs in highschool math were formulated.

Can't remeber if they were divergent or convergent series.

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