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Hi,

I was reading this when I came across Gourevitch's conjecture.

My understanding is that solutions to these series are of practical interest. If one encounters such a series, being able to solve it exactly is more practical than having to solve it numerically.

But, not being a mathematician, I simply can't imagine what the theoretical implications of proving such conjectures are.

What are they?

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3  
The theoretical implication is that this is really cool, and understanding why it is true is likely to require understanding something deep. –  Igor Rivin Aug 24 '12 at 20:05
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Isn't this like porn? –  Mariano Suárez-Alvarez Aug 24 '12 at 20:44
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Voted to close. As you seem to be an undergraduate in chemistry or related, it is unlikely we can convince you of the value of this, and I don't think anyone should try. If you find further questions, try math.stackexchange.com/questions –  Will Jagy Aug 24 '12 at 21:49
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@WillJagy How absurd, pejorative and close minded. Please tell me you're not serious. –  CHM Aug 25 '12 at 1:09
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CHM - Please don't be offended by the votes to close. This is not personal! Mathoverflow is specifically for questions that arise in mathematical research, as the FAQ makes clear. Your question is a great one, but as you admit is not a question that arose out of mathematical research. You might find that math.stackexchange.com would be a better venue. –  HJRW Aug 25 '12 at 7:47
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4 Answers

Let $k \ge 2$ be an ineteger. Consider the series $$ \sum_{n=1}^{\infty} \frac{1}{n^k} $$ this is typically denoted $\zeta(k)$ as it is the value of the Riemann zeta function at $k$.

Now, Euler showed that for even $k$ this is equl to $$q_k \pi^k $$ where $q_k$ is some rational number (that one can also describe explicty).

This is on the one hand an interesting fact and would also allow to calculate approximations of the powers of $\pi$ but what I actually want to get at is that from this it follows that if one knows that $\pi$ is transcendental then one gets directly that $\zeta(k)$ is transcendental and in particular not a rational number.

So this is for even $k$. What about odd $k$? Say $k=3$. Is this rational or irrational? This question was open for a long time until it was proved at the end of the 1970s by Apéry.

How does this prove go (very roughly!):

He first showed that $$ \zeta(3) = \frac{5}{2} \sum_{i=1}^{\infty} \frac{ (-1)^n}{n^3 C (2n , n )} $$ where $C(2n,n)$ is just the obvious binomial coefficient, which I fail to be able to type properly.

So one could say he evaluated the series on the right in 'a closed form'; showing that its values is something already known/defined.

Then based on this he derived some sequences of rational numbers that converge to $\zeta(3)$ so fast that it is impossible for $\zeta(3)$ to be rational itself thus proving the irrationality of $\zeta(3)$.

So, in order to show that $\zeta(3)$ is irrational he first needed to show that it is equal to the limit of this (other) series, or put differently to evaluate this series; not only the first simpler one.

It would be interesting to be able to do something like this for other odd numbers, but so far noone knows how to do so and the irrationality of $\zeta$ at any other odd positive integers is unknown (although there are results that assert that among certain collections of them there are at least some irrational ones).

Thus finding an evaluation of a series can be used to infer something theoretical on its value.

This is not always the motivation, but sometimes it is the case that the point is not so much to know the value of the series in order to replace it in some compuation say, but rather to use the series as a form of describing its value by simpler buildingblocks and thereby allowing to learn something (new) on the value.

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You could use(^{2n}_{n}) to get $(^{2n}_{n})$ –  Lunasaurus Rex Aug 25 '12 at 3:19
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To get $\binom{2n}{n}$ I've used \binom{2n}{n}. –  Américo Tavares Aug 25 '12 at 10:09
    
Thanks for the hints. (I will refrain from editing at least for the moent; but good to know for the future). –  quid Aug 26 '12 at 1:19
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Having the close form for a series of functions is also important:

  1. Closed forms can sometimes be used to find exact solutions to differential equations.
  2. Closed forms allow for accurate estimates on errors in approximations of special functions. Some of these are used in your calculator.
  3. Closed forms can sometimes produce proofs of non-series results. For example, the equation $$ e^{ix} = \cos(x) + i\sin(x) $$ is proven by using three Taylor series.
  4. This is a bit incestuous, but having the closed form for one series allows for manipulation to produce the closed form of other series.
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the identity in 3 is much more interestingly proven in other ways! :-) –  Mariano Suárez-Alvarez Aug 25 '12 at 4:56
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I can recommend reading "Closed forms: what they are and why we care" by Jon Borwein and Richard Crandall, the article is to appear in Notices Amer. Math. Soc. 60 (2013).

Edit (Dec 2012). The paper has just appeared in the Notices: pdf. Together with the sad news about Richard Crandall: he passed away.

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Wow. Anyone who's ever asked for a "closed form solution" should read this paper. Very eye-opening on a topic that I think most of us take for granted. –  Aeryk Aug 26 '12 at 2:08
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HERE is an example in MO. For the first question, I could evaluate the series in closed form, so I could compute with their known properties, and thus answer the question. For the second question, the series were just ${}_2F_1$ series, and I did not know their properties, so I could not answer the question. Despite the numerical evidence in the graphs.

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