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.

  • $\begingroup$ Which book${{}}$? $\endgroup$ – Lord Shark the Unknown May 5 at 16:19
  • $\begingroup$ @Lord Shark the Unknown the book is T.M. Apostol's Analytic Number Theory; page no. 310. $\endgroup$ – math maniac. May 5 at 16:20
  • $\begingroup$ That the LHS converges for $|x| < 1$ implies so does the RHS. Alternatively since $p(n) \ge 0$ that the RHS converges for $ x=r$ implies it converges for $|x| \le r$. Once both side converge the proof they are equal is the same for $x > 0$ or not. $\endgroup$ – reuns May 5 at 17:36
  • $\begingroup$ But before that we need to ensure whether LHS is indeed equal to RHS or not. The book only provides a rigorous proof of the above equality whenever $0 \leq x < 1.$ So my question is how do I extend this equality to the unit disk? $\endgroup$ – math maniac. May 6 at 17:02

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 $|z|<1$.

  • $\begingroup$ How do you get the inequality @Simply Beautiful Art $$|\log (1-z^m)| \leq |z|^m + \frac {|z|^{2m}} {|1-z^m|^2}$$? $\endgroup$ – A.Chattopadhyay May 13 at 5:38
  • $\begingroup$ From Taylor's theorem. $\endgroup$ – Simply Beautiful Art May 13 at 9:33

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