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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.

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  • $\begingroup$ You only need to evaluate the generating function at $t=-1$. $\endgroup$ Nov 8, 2019 at 14:41
  • $\begingroup$ @conditionalMethod Its not allowed to substitute the value of $t$ in the generating function $\endgroup$
    – rohit_r
    Nov 8, 2019 at 14:51
  • $\begingroup$ 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. $\endgroup$ Nov 8, 2019 at 14:56
  • $\begingroup$ @conditionalMethod The generating function is derived by assuming $t$ is very small. The cops are on their way! $\endgroup$
    – rohit_r
    Nov 8, 2019 at 15:06

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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 $x=-1$.

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