Geometric Progression and Counting Partitions

Question

Related with The number of partitions of $n$ into distinct parts equals the number of partitions of $n$ into odd parts

I know that ${(x^n-1)\over (x-1)}=1+x+x^2+x^3+x^4+...$ as a Geometric Progress, but I do not see how $${1\over (1-x)(1-x^3)(1-x^5)\cdot\dots}=(1+x+x^{1+1}+\dots)(1+x^3+x^{3+3}+\dots)(1+x^5+\dots).$$ can anyone please explain?

If you have the same linked question, then you could have added a comment on the answer to seek further clarification. If you have a specific question, go ahead and ask that, but fix the title of your post. Also learn MathJax. — StubbornAtom, Nov 04 '16 at 10:40

score 2 · Answer 1 · edited Nov 04 '16 at 11:32

2

Your question is why $1+x+x^2+x^3+x^4+...=\frac{1}{1-x}$

You know that ${x^n-1\over x-1}=1+x+x^2+x^3+x^4+...+x^{n-1}$

${(x^n-1)\over (x-1)}=\frac{x^n}{x-1}-\frac{1}{x-1}$

Let $n\to \infty$ and $-1<x<1$

$$\lim_{n\to \infty} \frac{x^n}{x-1}-\lim_{n\to \infty}\frac{1}{x-1}=0-\frac{1}{x-1}=\frac{1}{1-x}$$

If you substitute $x$ by $u^3$ or $u^5$ you´ll get similar results.

edited Nov 04 '16 at 11:32

Shagnik

3,663

answered Nov 04 '16 at 11:25

callculus42

31,012

OK so there is the implication that $$ lim_{n\to \infty}$$ and $$-1<x<1$$ – Artor Vrontayeus Nov 04 '16 at 13:00
@ArtorVrontayeus Yes, you are right. – callculus42 Nov 04 '16 at 13:02
Answer accepted with thanks. ( I do not know how else to give credit to callculus and Shagnic) – Artor Vrontayeus Nov 04 '16 at 13:14
@ArtorVrontayeus You only can accept the answer. Maybe you can vote up the answer as well. Shagnic does not get directly some points. But for a couple of edits Shagnic will receive some credits. – callculus42 Nov 04 '16 at 13:19

Geometric Progression and Counting Partitions

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