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In undergraduate courses we compute the sum $S$ of some series of the form $\frac{1}{P(n)}$ where $P(x)$ is some simple polynomial of degree $2$ with integer coefficients, by the following procedure:


(a) Choose an appropriate periodic function $f(x)$ defined over a domain $D.$

(b) Compute the Fourier series $S(x)$ of $f(x).$

(c) Choose a suitable $x$ in $D$ so that we obtain a linear equation for $S.$

(d) Solve the equation to get $S.$


When $P(x)=x^2+1$ we can take:

$f(x)= \exp(x),$ $D= [-\pi,\pi[$, and $x=\pi.$

$S$ is the sum from $n=1$ to infinity of $\frac{1}{n^2+1}.$

We get the equation:

$$ ch(\pi) = S(\pi) = 2\frac{sh(\pi)}{\pi}(\frac{1}{2}+S) $$ that gives $$ S=\frac{1}{2}(\frac{\pi}{th(\pi)}-1). $$

($ch,sh,th$ denote the classic hyperbolic functions)


Why this fails (in general) for polynomials $P(x)$ of degree $3.$ ?

Why this fails for the polynomial $P(x)=x^3.$ ?

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Sorry, but there is no way to understand the method from your post without knowing it beforehand. – darij grinberg Jan 5 '11 at 21:19
@darij: I am afraid... I thinked that the method is well known and everywhere used. Do not hesitate to edit my post if you feel this is appropriate to the understanding of the question. – Luis H Gallardo Jan 5 '11 at 21:29
Thing is, I don't know it. What is the relation between $P$ and the function $f$? – darij grinberg Jan 5 '11 at 21:40
And what does $S(\pi)$ mean? – darij grinberg Jan 5 '11 at 21:40
Oh, I see your first $S$ is different from your $S(x)$. Anyway, your comments are seriously messed up LaTex-wise, so I'm not any wiser noew. – darij grinberg Jan 6 '11 at 14:38

There is a systematic method for evaluating series of this type by residue calculus. It is explained in many texts on complex analysis. Using this method, certain sums over all integers can be evaluated. This makes use of functions like cotangent or cosecant, which have poles at all integers. An even function summed over the positive integers is easily reduced to a sum over all integers. For odd functions, this does not work. The method of summing series by residues is still applicable, but instead of the trig functions, we need to use a function that has poles only at integers of one sign. Such a function exists, namely the Gamma function, but it is not "elementary."

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Perhaps it fails because if it worked it would give an answer to a question that doesn't have one.

To be a little less cryptic, if there isn't any evaluation of the sum of the reciprocal cubes in terms of, say, the functions of 1st year calculus, then no method that yields only that kind of function is going to succeed in evaluating the sum.

And if you want to know why there should be no evaluation of that sum in those terms, well, when you find out, please let the rest of us know!

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@Gerry: ``James said suddenly: 'Well--I don't know, I can't tell'". John Galsworthy, The Forsyte Saga, page 430, line -2. – Luis H Gallardo Jan 6 '11 at 10:35
Misprint in formula of example corrected. – Luis H Gallardo Jan 8 '11 at 14:55
I believe that a shorter (but equivalent) answer would be: it would be nice to have a closed form for $\zeta(3).$ – Igor Rivin Jan 8 '11 at 16:44
@darij: Consider the $2\pi$ periodic function $f(x)$ such that $f(x)=x^2$ when $x \in [-\pi,\pi]$. Let $$ S(x) =\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n cos(nx)+b_nsin(nx). $$ be the Fourier series of $f$. Observe that: $b_n=0$ for all $n$. We get after some (short) computations: $$ a_0 =\frac{2 \pi^2}{3} $$ and for $n>0$ $$ a_n =\frac{4 \cdot (-1)^n}{n^2}. $$ Since $f(x)$ is continue over the reals, $S(x)=f(x)$ for all real numbers $x.$ Pick now $x=\pi.$ We get $$ \pi^2 = f(\pi)=S(\pi) = \frac{\pi^2}{3} +4S $$ Thus, $$ \zeta(2)=S= \frac{\pi^2}{6}. $$ – Luis H Gallardo Jan 9 '11 at 13:11

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