Theorem (Euler) $:$ For $|x|<1$ we have
$$\prod\limits_{m=1}^{\infty} \frac {1} {1-x^m} = \sum\limits_{n=0}^{\infty} p(n) x^n,$$ where $p(n)$ denotes the number of partitions of $n$ for $n \geq 1$ and $p(0)=1.$
In my book the above identity is proved only for $0 \leq x < 1$ and it was said at the end of the proof that this proof can be extended analytically to the whole unit disk. But how do I do that? Would anybody please help me regarding this?
What I have understood is that if I can show that $\lim\limits_{n \to \infty} \frac {p(n+1)} {p(n)} = 1$ then I am done. But I can't able to find the limit. Any help regarding this will be highly appreciated.
Thank you very much.
As mentioned by Lord Shark the Unknown, it is easy to prove this using the fact that uniformly converging analytic functions converge to analytic functions.
We have uniform convergence of an infinite product easily here on any disk of radius $r<1$ since
$$\left|\log\left[\frac1{1-z^m}\right]\right|=|\log(1-z^m)|\le|z|^m+\frac{|z|^{2m}}{|1-z^m|^2}\le r^m+\frac{r^{2m}}{1-r^m}$$
which converges by the ratio test for all $r\in[0,1)$.
Thus, the LHS is analytic on $|z|<1$.
Similarly for the RHS, we have the simple bound:
$$\left|\sum_{n=0}^\infty P(n)z^n\right|\le\sum_{n=0}^\infty P(n)|z|^n\le\sum_{n=0}^\infty P(n)r^n$$
The convergence of the last sum is also very likely proven in your book for $r\in[0,1)$.
Thus, the RHS is analytic on $|z|<1$.
Thus it follows from a simple analytic continuation argument that this holds for all $|z|<1$.