# Summation of Legendre polynomial series

How do I find the sum $$\sum_{n=0}^{\infty}(-1)^n P_n(x)$$ where $$P_n$$ are the $$n$$th order Legendre polynomials? I tried using the generating function but I was not able to arrive at an answer. Any hints appreciated.

• You only need to evaluate the generating function at $t=-1$. Nov 8, 2019 at 14:41
• @conditionalMethod Its not allowed to substitute the value of $t$ in the generating function Nov 8, 2019 at 14:51
• I don't know who will stop me. Oops, I just did it. Wait, I will even dare to write $\sum_{n=0}^{\infty}(-1)^nP_n(x)=\frac{1}{\sqrt{2+2x}}$. Send the cops. Nov 8, 2019 at 14:56
• @conditionalMethod The generating function is derived by assuming $t$ is very small. The cops are on their way! Nov 8, 2019 at 15:06

I don't see any problem in using the generating function. $$\frac{1}{\sqrt{r^2-2rx+1}} = \sum_{n=0}^{\infty} P_n(x)r^n$$ holds for all $$|r|<1$$ (as the radius of convergence in the complex plane is the distance to the first singular point, which here lie both on the unit circle) and $$|x| \le 1$$ and it holds also for $$r=-1$$, your example. Thus $$\sum_{n=0}^{\infty} P_n(x)(-1)^n = \frac{1}{\sqrt{2+2x}}$$ As already remarked by $${conditionalMethod}$$. Note that this leads to an ambigious result for $$x=-1$$.