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Recently I came across this paradoxical equation:

$$1-2+3-4+\dots = \frac{1}{4}$$

More rigorously, this infinite sum can be write as

$$\sum_{n=1}^{\infty} n(-1)^{n-1}$$

and it is well known to diverge since, for instance

$$1=1$$

$$1-2=-1$$

$$1-2+3=2$$

$$1-2+3-4=-2$$

$$1-2+3-4+5=3$$

$$1-2+3-4+5-6=-3$$

Now, the thing I cannot explain to myself is why the sum equals $\frac{1}{4}$, although it doesn't tend towards any limit. Here's because, I think, the equality above is called "paradoxical".

I (tried to) read the Wikipedia article, but it is full of material I've never encountered in my studies, full of examples and analogies that made me even more confused.

The only thing I believe to have understood concerning this equality, written for the first time by Leonhard Euler, is that the latter came to the rusult by reducing the polynomial $$1-2x+3x^2-4x^3+\dots$$ in $$\frac{1}{(1+x)^2}$$

Thus, if we take $x=1$, we obtain the equality above. But how is this reduction done?

I'm afraid if I couldn't realise the answers to this question on Wikipedia, but I just need an explanation that would make me understand, even intuitively, the paradoxical nature of the equality and its mathematical derivation by means of even one clear reasoning.

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    $\begingroup$ You can't just plug any value into the power series, or you'd also get $1+x+x^2+\dots = \frac{1}{1-x}$ for all $x$, which would have values $x>1$ converging to negative values. There is no paradox here, you are just mis-using limits. $\endgroup$ – Thomas Andrews Aug 5 '13 at 16:24
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    $\begingroup$ @ThomasAndrews: Not just any value, but a value on the circle of convergence, in the case of Abel summability. $\endgroup$ – Robert Israel Aug 5 '13 at 16:43
  • $\begingroup$ I wrote an essay about this in college. I've posted it on my blog, but it's pretty dry writing... Still, you might find it useful. $\endgroup$ – BlueRaja - Danny Pflughoeft Aug 5 '13 at 20:23
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    $\begingroup$ If you are interested in the topic, I suggest watching some of Carl Bender's lectures here. $\endgroup$ – Ali Aug 6 '13 at 1:21
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    $\begingroup$ @Ali Thank you very much, it's very interesting! :) $\endgroup$ – Riccardo Del Monte Aug 6 '13 at 7:49
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You have to be very careful with power series. They all have a range of values for which they work.

Consider the power series:

$$P = 1 - x + x^2 - x^3 + x^4 - x^5 + x^6 \pm \cdots $$

This is a geometric series with first term $a=1$ and common ratio $r=-x$. For an infinite geometric series, if $-1 < r < 1$ then it converges to a finite value given by a well-known formula: $$P = \frac{1}{1+x}$$

It's tempting to think that the power series is equal to this fraction, i.e. $$\frac{1}{1+x} \equiv 1 - x + x^2 - x^3 + x^4 - x^5 + x^6 \pm \cdots $$ However, this is only true when $-1 < x < 1$. For any value of $x$ between $-1$ and $1$, the right-hand side will give the same value as the left-hand side. You can put in values outside of that rangle, but you will get crazy answers. For example, try $x=-2$. We get $$-1 = 1 + 2 + 4 + 8 + 16 + 32 + \cdots $$ The expression that you gave is found by differentiating both sides of $$\frac{1}{1+x} \equiv 1 - x + x^2 - x^3 + x^4 - x^5 + x^6 \pm \cdots $$ The quotient rule on the LHS, the power rule to the RHS and then multiplying both sides by $-1$ gives $$\frac{1}{(1+x)^2} \equiv 1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 \pm \cdots $$ Again, this is only valid for values of $x$ between $-1$ and $1$. In fact, you can let $x$ be a complex number and insist that $|x|<1$. If you put things like $x=1$ then you will get nonsense because $x=1$ does not satisfy $-1<x<1$.

Some people to choose to assign a value to these series by allowing any values of $x$ whatsoever. This is related to the idea of analytic continuation.

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  • $\begingroup$ Just one thing: Is $1-2x+3x^2-4x^3+\dots$ a geometric series or not? It doesn't seem to have a constant ratio between successive terms. So, if it is not, how can you say when it's convegent and when it's divergent? Is $-1<x<1$ still valid in this case? $\endgroup$ – Riccardo Del Monte Aug 6 '13 at 8:06
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    $\begingroup$ @RiccardoDelMonte No, it isn't a geometric series. It's just a power series. Every power series has a radius of convergence and in this case $\rho = 1$, which means it converges for all $|x| < 1$. In the real case that means $-1 < x < 1$. en.wikipedia.org/wiki/Radius_of_convergence $\endgroup$ – Fly by Night Aug 6 '13 at 12:05
  • $\begingroup$ Ok, got it. Thanks $\endgroup$ – Riccardo Del Monte Aug 6 '13 at 15:15
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This is known as Analytic Continuation. It is done exactly as you explained - you have a function that converges on some interval and then you extend the domain of the function past the interval it converges on by using some closed result, in this case $ \frac{1}{(1 + x)^2} $.

Another good example is $ 0! - 1! + 2! - 3! + \cdots $ although this one's closed form is a bit more involved.

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    $\begingroup$ That's probably not a good example to use here, since the series $\sum_{k} k! x^k$ has radius of convergence $0$. You need other tricks to sum it, e.g. Borel summability. $\endgroup$ – Robert Israel Aug 5 '13 at 16:33
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The other answers are good as they more or less explain what is going on. This answer will seek to explain the $\frac{1}{4}$. Let us first look at a simpler series that exhibits similar enough behavior. That is the series, $$1-1+1-1+1-1\cdots.$$ Note that this is a divergent geometric series, but it is a a fly's hair away from being convergent. What do I mean? We can write down a power series, $$\frac{1}{1+x}=1-x+x^x-x^3+\cdots\mbox{ when } |x|<1.$$ If we blindly plug in $x=1$ (even though we should not be allowed to) then we get the series obove is equal to $\frac{1}{2}$. Now why is $\frac{1}{2}$ a sensible answer? If we look at the partial sums, we get the series, $$a_n=\frac{1}{2}(1+(-1)^n)$$. Now if we take the means of the partial sums, of $a_n$ we get $$\lim_{n\to\infty}\frac{1}{n+1}\Sigma_0^n a_n=\frac{1}{2}.$$ This is a compromise between the two points we are jumping between. However, this was not you question, but this should be a sort of warm up, so let us try to do something similar with your series and see if we can extract $\frac{1}{4}$. We have the series $$1-2+3-4+5-6+\cdots.$$ If we look at the partial sums as before, we get $$1,-1, 2, -2, 3, -3, 4,-4,\cdots.$$ Now let us take the means of the partial sums of this new sequence. This gives us $$1,0,\frac{2}{3},0,\frac{3}{5},0,\frac{5}{7},0\cdots.$$ Note that this is the result of two sub-sequences weaved together. The firts is the sub-sequence defined by $$\frac{n}{2n+1}$$, and the other is the constant zero sequence. The sequence $$\frac{n}{2n+1}$$ has limit $\frac{1}{2}$ while the constant zero sequence has limit zero. Now if we compromise between the two subsequences we get the desired $\frac{1}{4}$. This is related to Cesaro Summation. In fact this is just an iterated version.

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We need to distinguish between different ways to boil down a series to a single number. The sum of a series is the limit of the partial sums in the topology of the real numbers. The Abel sum of a series is, roughly speaking, the limit of $\sum a_nx^n$ as $x\to 1$.

These two procedures are closely related. Abel's theorem states that if the sum of a series exists, then the Abel sum also exists and is the same number. However, the converse is not true!

The series $1-2+3-4+\cdots$ is an example where the Abel sum exists but the sum does not exist. This result is paradoxical only if you think that the sum and the Abel sum are the same thing. They aren't, and that resolves the paradox!

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  • $\begingroup$ Thank you, especially for the difference between the sum and the Abel sum of a series. Now I have clearer ideas. :) $\endgroup$ – Riccardo Del Monte Aug 5 '13 at 19:39

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