How one could prove, that q pochhammer symbol $(1,1/n) = \prod_{k = 1}^{\infty}(1-\frac{1}{n^k}) \geq 1 - \frac{1}{n-1}$
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2 Answers
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$$\prod (1-x_i)=1-x_1-x_2(1-x_1)-x_3(1-x_1)(1-x_2)-\ldots \geqslant 1-\sum x_i, \forall x_i\in [0,1],$$ use this for $x_i=1/n^i$.
answered Apr 7, 2019 at 11:32
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This function has an interesting asymptotic expansion. $$\prod_{k\ge1}(1-n^{-k}) = 1 - n^{-1} - n^{-2} + n^{-5} + n^{-7} - n^{-12} - n^{-15} + \cdots\,\,. $$ The coefficients are all $\pm 1$ and the exponents are generalized pentagonal numbers.
answered Apr 7, 2019 at 14:05
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$\begingroup$ Ah, by the way it also implies the inequality asked by OP! I remember that there was a question on MO with tremendously hard proofs of simple facts, this argument may be included in the collection. $\endgroup$ Commented Apr 7, 2019 at 20:57