# Curious identities

1. May 6, 2010

### Dickfore

These identities might be considered as true:

$$\begin{array}{rcl} 1 - 1 + 1 - 1 + \cdots & = & \frac{1}{2} \\ 1 - 2 + 3 - 4 + \cdots & = & \frac{1}{4} \\ 1 - 4 + 9 - 16 + \cdots & = & 0 \\ 1 + 1 + 1 + 1 + \cdots & = & -\frac{1}{2} \\ 1 + 2 + 3 + 4 + \cdots & = & -\frac{1}{12} \\ 1 + 4 + 9 + 16 + \cdots & = & 0 \end{array}$$

2. May 7, 2010

### mathman

They might not.

3. May 7, 2010

### Martin Rattigan

This identity may be considered as true:

4. May 7, 2010

### arildno

Not with the summation operation as ordinarily defined.

Then, they are totally untrue.

5. May 7, 2010

### Count Iblis

$$\sum_{n=0}^{\infty}(-1)^{n} =\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!} n! = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!} \int_{0}^{\infty}x^{n}\exp(-x)dx$$

If we interchange the order of summation and integration we get a finite expression. This is known as Borel resummation. Of course, that's not a valid operation in a narrow minded sense, but then the original summation was presumably the result of an illegal formal operation to begin with. Because we are only performing formal operations, the information present in the original mathematical expression will not get lost. If that was equivalent to some number, we're likely to recover precisely that number.

So, we get:

$$\int_{0}^{\infty}\exp(-2x)dx = \frac{1}{2}$$

Similarly:

$$\sum_{n=0}^{\infty}(-1)^{n+1}n = \sum_{n=0}^{\infty}\frac{(-1)^{n+1}n}{n!} \int_{0}^{\infty}x^{n}\exp(-x)dx$$

Interchanging integration and summation gives:

$$\int_{0}^{\infty}x\exp(-2x)dx =\frac{1}{4}$$

6. May 7, 2010

7. May 7, 2010

### Martin Rattigan

Remarkably well spotted Count Iblis!

Your entry suggests the Borel resummation could be useful in some circumstances - do you have an example?

Last edited: May 7, 2010
8. May 7, 2010

9. May 7, 2010

### Dickfore

These "tricks" are used in Quantum Field Theory. See Zeta function regularization.

10. May 8, 2010

### fluidistic

Sorry I've no idea how $$\sum _{i=1}^{\infty} i = -\frac{1}{2}$$. In fact your 3 last identites make absolutely no sense to me. I've checked out http://en.wikipedia.org/wiki/Zeta_function_regularization and I really don't see how can these sums be true. It seems you have used the Cesaro's sum for the first series. Probably with the 3 firsts. But the 3 lasts... what the heck?

11. May 8, 2010

### Dickfore

Let

$$x_{k} \equiv \sum_{n = 1}^{\infty}{n^{k}} = 1^{k} + 2^{k} + 3^{k} + 4^{k} + \cdots$$

and

$$y_{k} \equiv \sum_{n = 1}^{\infty}{(-1)^{n - 1} n^{k}} = 1^{k} - 2^{k} + 3^{k} - 4^{k} + \cdots$$

Then, we can formally write:

$$x_{k} = y_{k} + 2 \sum_{n = 1}^{\infty}{(2 n)^{k}} = y_{k} + 2^{k + 1} x_{k}$$

$$x_{k} = \frac{y_{k}}{1 - 2^{k+1}}$$

If we accept the first three identities as correct, then the last three follow for $k = 0, 1, 2$.

12. May 8, 2010

Knopp, in his (now classic) book on infinite series, introduces these ideas in Chapter XIII. In the introductory pages of that chapter, he writes that Euler used the following (quoting here, in my copy this is from pags 457-458).

in our exposition, the symbol for infinite sequences was created and then worked with; it was not so originally, these sequences were there, and the question was what could be done with them.

On this account, problems of convergence in the modern sense were at first remote from the minds of the mathematicians. Thus it is not to be wondered at that Euler, for instance, uses the geometric series

$$1 + x + x^2 + \dots = \frac 1{1-x}$$

even for $x = -1$ or $x = -2$, so that he unhesitatingly wrote

$$1 -1 + 1 - 1 + \dots = \frac 1 2$$

or

$$1 - 2 + 2^2 - 2^3 + \dots = \frac 1 3$$

Similarly, from

$$\left(\frac 1 {1-x} \right)^2 = 1 + 2x + 3x^2 + \dots$$

he deduced the relation

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

and a great deal more. It is true that most mathematicians of those times held themselves aloof from such results in instinctive mistrust, and recognized only those that are true in the modern sense. But they had no clear insight into the reasons why one type of result should be admitted and not the other.

***********A short jump in quote - the following is still on page 458.

It is clear that the convention has no precise basis. Even though, for instance, the series $1 - 1 + 1 - 1 + \dots$ results in a very simple manner from the division ${1}/{(1-x)}$ for $x = -1$ (see above), and thus should be equated to $\frac 1 2$, there is no reason why the same series should not result from quite different expressions and why, in view of those other methods of obtaining it, it should not be given a different value. The above series may actually be obtained, for $x = 0$, from the function $f(x)$ represented for every $x > 0$ by the Dirichelet series

$$f(x) = \sum_{n=1}^\infty {\frac{(-1)^{n-1}}{n^x}} = 1 - \frac 1 {2^x} + \frac 1 {3^x} - \frac 1 {4^x} + \dots$$

or from

$$\frac{1+x}{1 +x + x^2} = \frac{1-x^2}{1-x^3} = 1 - x^2 + x^3 - x^5 + x^6 - x^8 + \dots$$

by putting $x = 1$. In view of this latter method of deduction, we should have to take $1 - 1 + \dots = \frac 2 3$, and in the case of the former there is no immediate evidence what value $f(0)$ may have; it need not at any rate be $+ \frac 1 2$.

******************End of quotations

Knopp uses this discussion as a launching platform for his introduction and discussion of various issues of summation, divergent series, asymptotic series, and applications. It is an interesting, and enlightening, read.

13. May 8, 2010

### Count Iblis

Last edited by a moderator: May 4, 2017
14. May 8, 2010

### Martin Rattigan

Well, for my part thanks for the info. Interesting!

15. May 9, 2010

### The Chaz

I almost woke my kid up with my laughter!

16. May 31, 2010

### Count Iblis

This is also nice:

$$\prod_{k=1}^{\infty}k=\frac{1}{\sqrt{2\pi}}$$

17. Jun 1, 2010

### NaturePaper

Then what is x_k (I don't think it is a finite number) and then what is meant by your last derivation? :)

18. Jun 2, 2010

### ice109

most of these identities are true if you replace regular summation with some sort of other summation like cesaro summation or abel summation or borel summation.

19. Jun 2, 2010

### Dickfore

Take the logarithm of both sides:

$$\sum_{n = 1}^{\infty}{\ln{n}} = -\frac{\ln(2 \pi)}{2}$$

The sum can be related to the derivative of the Riemann zeta function:

$$\zeta(s) = \sum_{n = 1}^{\infty}{\frac{1}{n^{s}}}$$

$$\zeta'(s) = \sum_{n = 1}^{\infty}{-\ln{n} \, e^{-s \, \ln{n}}}$$

$$\sum_{n = 1}^{\infty}{\ln{n}} = -\zeta'(0)$$

So, your statement is equivalent to saying:

$$\zeta'(0) = \frac{1}{2} \, \ln(2 \pi)$$

20. Jun 2, 2010

### Count Iblis

Dickfore, from your derivation, I now see that I forgot a minus sign. The correct expression is:

$$\prod_{k=1}^{\infty}k = \sqrt{2\pi}$$