Infinite sums of inverses of degree $3$ polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:01:36Z http://mathoverflow.net/feeds/question/51246 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51246/infinite-sums-of-inverses-of-degree-3-polynomials Infinite sums of inverses of degree $3$ polynomials Luis H Gallardo 2011-01-05T20:56:47Z 2011-02-20T14:22:15Z <p>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:</p> <p>(sketch)</p> <p>(a) Choose an appropriate periodic function $f(x)$ defined over a domain $D.$</p> <p>(b) Compute the Fourier series $S(x)$ of $f(x).$</p> <p>(c) Choose a suitable $x$ in $D$ so that we obtain a linear equation for $S.$</p> <p>(d) Solve the equation to get $S.$</p> <p>Example: </p> <p>When $P(x)=x^2+1$ we can take:</p> <p>$f(x)= \exp(x),$ $D= [-\pi,\pi[$, and $x=\pi.$ </p> <p>$S$ is the sum from $n=1$ to infinity of $\frac{1}{n^2+1}.$</p> <p>We get the equation: </p> <p>$$ch(\pi) = S(\pi) = 2\frac{sh(\pi)}{\pi}(\frac{1}{2}+S)$$ that gives $$S=\frac{1}{2}(\frac{\pi}{th(\pi)}-1).$$</p> <p>($ch,sh,th$ denote the classic hyperbolic functions)</p> <p>Question:</p> <p>Why this fails (in general) for polynomials $P(x)$ of degree $3.$ ? </p> <p>Why this fails for the polynomial $P(x)=x^3.$ ?</p> http://mathoverflow.net/questions/51246/infinite-sums-of-inverses-of-degree-3-polynomials/51278#51278 Answer by Gerry Myerson for Infinite sums of inverses of degree $3$ polynomials Gerry Myerson 2011-01-06T04:12:33Z 2011-01-06T04:12:33Z <p>Perhaps it fails because if it worked it would give an answer to a question that doesn't have one. </p> <p>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. </p> <p>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!</p> http://mathoverflow.net/questions/51246/infinite-sums-of-inverses-of-degree-3-polynomials/51543#51543 Answer by Michael Renardy for Infinite sums of inverses of degree $3$ polynomials Michael Renardy 2011-01-09T14:10:11Z 2011-01-09T14:10:11Z <p>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."</p>