I find it difficult to get my head around the idea that an infinite series of rational terms can yield an irrational result, e.g. 1+1/2^2+1/3^2+1/4^2+...=pi^2/6. There are certainly infinite series of rational terms that yield a rational result.

Is there a test that can be performed on series that enables you to determine whether the series has a rational or irrational sum even if you can't determine what the sum is?

Thanks

Jeff

In general, no. There are exceptions; for example mathematicians have proven that 1/1³ +1/2³ +1/3³ +... is irrational, and it isn't very hard to prove that 1/1 + 1/(1x3)+1/(1x3x5)+... is irrational (once you've seen the proof for e's irrationality), but by and large, it is not known whether most series are rational or irrational, and there isn't a test to apply, at least yet...

The reason is simply because there are so many varieties of series, and proving series irrational is at a very high level of mathematics, combining analysis with number theory. It currently isn't even known whether a series as simple as 1/1² -1/3² +1/5² -1/7² +... is irrational, although it's strongly suspected to be.

Brad

By the way, you mentioned that you couldn't quite see how an infinite series of rational terms can become irrational. This is largely because ordinary intuition tends to fail when dealing with the infinite. And, this is of course because intuition is largely, if not fully, based in experience, which we have none of in the infinite. I'll readily admit the idea is odd though. Just as an example of why this could occur, consider

.1010010001...=10^-1 + 10^-3 + 10^-6 + 10^-10 + ...

which obviously doesn't have a recurring decimal pattern, and therefore isn't rational. If you'd like, I could also prove the irrationality of e here, it's not too difficult, and if you haven't seen it before, you'll probably find it interesting.

Brad

I would like to see the proof, if you don't mind...

Thanks,
Yatir

Kerwin

I don't undestande what you mean by the first term.
I don't really know much about this subject.
So could you please explain step-by-step what you've done.

Thanks,
Yatir

Yatir, suppose that e is equal to some fraction p/q. Then

e = p/q = 1/0! + 1/1! + 1/2! + 1/3! + ... + 1/q! + 1/(q+1)! + ....

Multiplying by q!, and noting that q!/x! is integer as long as x is less than or equal to q, (and letting I₁ , and I₂ be arbitrary integers)

(q-1)!p = I₁ = (q!/0!+ q!/1! + ... q!/q!) + 1/(q+1) + 1/[(q+1)(q+2)] + ...
= I₂ + 1/(q+1) + 1/[(q+1)(q+2)] + ...

Therefore, we know that 1/(q+1) + 1/[(q+1)(q+2)] + ... is some integer, and it's obvious that it is greater than 0. But,

1/(q+1) + 1/[(q+1)(q+2)] + ... < 1/(q+1) + 1/(q+1)² + 1/(q+1)³ + ... = 1/q £ 1

Therefore, there is an integer between 0 and 1, which is a contradiction.

Brad

Yes. Thank you I see it now.
I remember once seeing another proof of that, it was quite complicated. This is rather a very simple proof.
I hate to trouble you so much, but how do we know that: 1/(q+1)+1/(q+1)^2+...=1/q ?

Thanks,
Yatir

Geometric progression

Result: Sum to infinity of Geometric progression when |r|< 1 :

S = a/(1-r)

where a is first term and r is common ratio

Tell me if you have seen the proof for the result...

George

Oh! Ofcourse.
I haven't noticed that.

Thnx,

Yatir

Thanks Brad for the reply. You do add an intriguing 'not yet...'. Does that mean that people believe that it may be possible to devise a test? Do any tests exist even for a limited class of series?

Jeff

It seems unlikely that any test could be devised, simply because there are such a wide variety of series. Of course, I don't have a proof of this, so we can't be sure there is no such test, but it seems unlikely. There are, though, a couple of "tests" that can be used quite a bit. They're really not tests, however, just tricks for a proof.

The function f(x)=xⁿ (1-x)ⁿ /n! crops up quite a bit in proofs, namely the irrationality proof of pi and of e^y for any non-zero rational y. I can elaborate on this if you'd like.

A second method uses some of the time is called irrationality measure. I'd unfortunately be hard pressed to explain this very well as I've just learned it myself, but basically, we derive a statement that some number minus any rational number must have a difference between them. For example, it can be shown that

Therefore, $\pi$ can never be rational, or else we'd have $0>1/q^{24}>0$. I'm pretty sure this method isn't used as much as it would seem though; more often it's used to find out how far a rational approximation for a number is off, which is useful in computation.

Brad