I'm reading about Legendre polynomials for additional information since it is interesting to know! Moreover it would help me with
a task I am working on. See
<https://math.stackexchange.com/questions/3945490>


The generating function of Legendre polynomials $P\_n(x)$ is defined as
$$\frac{1}{\sqrt{1-2tx+t^2}}=\sum\limits\_{n=0}^{\infty}P\_n(x)t^n.$$


It is known that for large $n$ the asymptotic expansion of the Legendre polynomials is


$$P\_n(x) \sim \frac{1}{\sqrt{2\pi n}}\frac{(x+(x^2-1)^{1/2})^{n+1/2}}{(x^2-1)^{1/4}}.$$


My question is how to prove this, without using integrals (if there exists such a proof). I searched a lot, but all the proofs I found regarding this start from the integral representation of the Legendre polynomials. Is it possible to prove it by starting directly from the generating function?


Thanks!



How to obtain the asymptotics of Legendre polynomials directly from their generating function
You may write $2x=a+1/a$ for certain $a$, $|a|>1$ (I guess you mean $|x|>1$), then
$$
\frac1{\sqrt{1-2tx+t^2}}=
\frac1{\sqrt{(1-at)(1-a^{-1}t)}}\\= \sum (-1)^n{-1/2\choose n}a^nt^n\cdot \sum (-1)^n{-1/2\choose n}a^{-n}t^n\\
:=\sum c\_na^nt^n\cdot \sum c\_na^{-n}t^n,\quad c\_n=(-1)^n{-1/2\choose n}=\frac{1\cdot 3\cdot\ldots \cdot(2n-1)}{2\cdot 4\cdot \ldots \cdot (2n)}\sim \frac1{\sqrt{\pi n}}
$$
(see the last relation in "Additional identities" section [here](https://en.wikipedia.org/wiki/Double_factorial), or deduce from the [Wallis formula](https://en.wikipedia.org/wiki/Wallis_product)).
Thus $$P\_n(x)=\sum\_{k=0}^n c\_{n-k}c\_k a^{n-2k}.$$
This gives (only small values of $k$ matter) the asymptotics $$P\_n(x)a^{-n}\sqrt{\pi n}\to\_{(\*)} \sum\_{k=0}^\infty c\_k a^{-2k}=\frac1{\sqrt{1-a^{-2}}}$$
that is equivalent to what you ask for.


More details for $(\*)$. We have
$$
P\_n(x)a^{-n}\sqrt{\pi n}-\sum\_{k=0}^\infty c\_k a^{-2k}=\sum \alpha\_kc\_ka^{-2k},
$$
where
$$\alpha\_k=\begin{cases}\sqrt{\pi n}c\_{n-k}-1,& k\leqslant n\\
-1,&k>n.\end{cases}$$
For any fixed $k$ we have $\alpha\_k\to 0$ when $n\to \infty$. On the other hand, $-1\leqslant \alpha\_k\leqslant 100k$, say (for $k\geqslant n/2$ use the bound $c\_{n-k}\leqslant 1$; for $k<n/2$ use the bound $c\_{n-k}\leqslant c\_{\lceil n/2\rceil}\leqslant 1/\sqrt{\pi n/2}$), thus the series $\sum \alpha\_k c\_k a^{-2k}$ is dominated by an absolutely convergent series $\sum 100 kc\_ka^{-2k}$ and its sum goes to 0 by Dominated Convergence Theorem.)


