0

For $1\gg \varepsilon > 0$, does this following sequence converge to $0$? $$ a_N = \Pi_{n=2}^N \left(1-\frac{1}{n^{1+\varepsilon}}\right) $$ This is interesting, because if $\varepsilon = 0$, the product telescopes: $$ \Pi_{n=2}^N \left(1-\frac{1}{n}\right) = \frac{1}{2} \times \dots \times \frac{N-1}{N} = \frac{1}{N} \to 0. $$ I'm wondering if we weaken the term $\frac{1}{n}$ by adding an $\varepsilon$ power, we still get convergence to $0$. If we let $\varepsilon = 1$, then we don't get this property. $$ \Pi_{n=2}^N \left(1-\frac{1}{n^2}\right) = \frac{(1)(3)}{2^2} \times \dots \times \frac{(N-1)(N+1)}{N^2} \propto_{N} \frac{N-1}{N} \to 1. $$

Edit: A comment indicates that there exists a relevant theorem. If so, I'd appreciate an answer or comment that links to or proves this theorem (just to verify that the theorem is correct and to get some interesting ideas out of the proof).

Albert
  • 1,609
  • 4
    No, there is a theorem, for positive sequence $b_n<1$ that $\prod(1-b_n)$ converges to a non-zero value iff $\sum b_n$ converges. So your product doesn't diverge for $\epsilon>0.$ – Thomas Andrews Apr 04 '24 at 02:42
  • @ThomasAndrews Would you have a source for this theorem? – Albert Apr 04 '24 at 02:44
  • I've forgotten the name of the theorem, if it has a name, so hard to look it up, but it isn't hard to use various inequalities about the logarithm to prove this, I believe. – Thomas Andrews Apr 04 '24 at 02:54
  • Edit: A comment by @ThomasAndrews indicates that there exists a relevant theorem. If so, I'd appreciate an answer or comment that links to or proves this theorem (just to verify that the theorem is correct and to get some interesting ideas out of the proof). – Albert Apr 04 '24 at 03:00
  • Put your edit in the question, not in a comment. A reader shouldn't have to read all the comments to know your question. – Thomas Andrews Apr 04 '24 at 03:01
  • The proper proof is to show the lemma $$\log(1-b)\geq - \frac{b}{1-b},$$ for $0<b<1.$ Then show $\frac 1{1-b_i}$ has a real upper bound if $\sum b_i$ converges and all $b_i\neq1.$ – Thomas Andrews Apr 04 '24 at 03:13
  • $$ a_N = \exp \left( {\sum\limits_{n = 1}^N {\log \left( {1 - \frac{1}{{n^{1 + \varepsilon } }}} \right)} } \right) \le \exp \left( { - \sum\limits_{n = 1}^N {\frac{1}{{n^{1 + \varepsilon } }}} } \right) \le \exp \left( { - \sum\limits_{n = 1}^\infty {\frac{1}{{n^{1 + \varepsilon } }}} } \right) < + \infty $$ since $\log(1-x)<-x$ for $0<x<1$. – Gary Apr 04 '24 at 04:57
  • See https://math.stackexchange.com/questions/1740706/proving-a-result-in-infinite-products-prod-1a-n-converges-to-a-non-zero?rq=1 for the theorem that Thomas Andrews mentioned. It's also mentioned on the Wiki page of infinite product – ioveri Apr 04 '24 at 08:21

0 Answers0