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If I do:

 NSum[(i + 1)/(i + 2) LegendreP[i, 0] LegendreP[i, 0], {i, 0, Infinity}]

I get:

1.25216

If I do:

Sum[(i + 1)/(i + 2) LegendreP[i, 0] LegendreP[i, 0], {i, 0, 1000}] // N

The result is:

2.40862

This last result is larger than the infinite sum, but the sum terms are all nonnegative. That is not possible. Am I wrong or is Mathematica wrong?

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  • $\begingroup$ The odd terms are zero. $\endgroup$ Commented Aug 28, 2013 at 17:29
  • $\begingroup$ One thing is that I am not sure if the sum diverges. If this is the case, the infinity sum is wrong anyway. $\endgroup$ Commented Aug 28, 2013 at 17:53
  • $\begingroup$ The sum diverges. $\endgroup$ Commented Aug 28, 2013 at 18:14
  • $\begingroup$ But see here SumConvergence[(i + 1)/(i + 2) LegendreP[i, 0] LegendreP[i, 0], i, Assumptions -> Element[i, Integers] && Mod[i, 2] == 0]! $\endgroup$ Commented Aug 28, 2013 at 18:17
  • $\begingroup$ Interesting. I did an Integral test and it diverged. $\endgroup$ Commented Aug 28, 2013 at 18:38

1 Answer 1

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We have the known result (see e.g. Abramowitz and Stegun)

$$P_{2n}(0)=\left(-\frac14\right)^n\binom{2n}{n}$$

Substituting this result into your sum (while also exploiting the oddness of the odd-order Legendre polynomials) yields

$$\sum_{k=0}^\infty\frac{2k+1}{2k+2}\binom{2k}{k}^2\left(\frac1{16}\right)^k$$

which Mathematica says is divergent, and that is certainly because the central binomial coefficients grow quite quickly ($\binom{2k}{k}\sim\frac{4^k}{\sqrt{\pi k}}$).

Nevertheless, we can ask Mathematica to evaluate the sum

$$\sum_{k=0}^\infty\frac{2k+1}{2k+2}\binom{2k}{k}^2 z^k$$

Sum[(((2 k + 1) Binomial[2 k, k]^2)/(2 k + 2)) x^k, {k, 0, ∞}] // FullSimplify
   (EllipticK[16 x] - EllipticE[16 x])/(8 π x)

where we have obtained a result in terms of complete elliptic integrals.

Unfortunately, it is also known that $K(m)$ exhibits a logarithmic singularity at $m=1$, so we again reach the conclusion that the original sum is indeed divergent.

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