I am a physicist working on an electrostatic problem and this series popped up: $\sum^{\infty}_{l=0} (P_l(x))^2$ where $P_l$ is the $l$-th Legendre polynomial. Computing this numerically I think the series converges for $x\in(-1,1)$. I don't have the proper knowledge and experience to find out whether it really does. So: 1. Does the series converge? 2. If it does - what is the sum?

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Legendre_polynomials $\endgroup$ Mar 12, 2012 at 9:32
  • $\begingroup$ Thanks, but I still can't see how to prove it converges. I have a feeling it could be proven with the ratio test, however I don't see the solution. And I have no clue at all about the sum. $\endgroup$
    – Jože
    Mar 12, 2012 at 10:19
  • $\begingroup$ Although it does not converge, I would think there might be an associated distribution for it, like dirac delta or its derivatives... See www-elsa.physik.uni-bonn.de/~dieckman/InfProd/… $\endgroup$
    – Machinato
    Jun 15, 2018 at 14:24

1 Answer 1


This question is delicate, but the series does not always converge. A formula of Laplace ((8.21.2) in Szego's book Orthogonal Polynomials) states that for $0<\theta<\pi$, $$ P_n(\cos\theta) = 2^{1/2}(n\pi\sin\theta)^{-1/2}\cos((n+\frac12\theta)-\pi/4) + O(n^{-3/2}).$$ Consequently $\sum |P_n(\cos \theta)|^2$ converges if and only if $$\sum n^{-1} \cos^2\left((n+\frac12)\theta-\pi/4\right)$$ does. But this series does not converge for many values of $\theta$ rationally related to $\pi$, for which there is an arithmetic progression of terms equal to $1/n$. The answer to this question may be relevant when $\theta$ is not rationally related to $\pi$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .