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The product has only positive factors so it has zero as lower bound. Also the product is decreasing as all its factors are less than one. In conclusion the series must have a limit. I also compute the first 150 values of the product and I got around 0.297. I believe that the product converges very, very, slowly to zero, but I can't prove it.

Simply Beautiful Art
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motoras
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  • It is bounded by zero (positive) and continuously decreasing, so it should converge to zero, no? – user1952500 Aug 17 '17 at 01:16
  • @user1952500 so is the sequence $a_n = (n + 1)/n$ – Trevor Gunn Aug 17 '17 at 01:17
  • @TrevorGunn, do you mean the $\sum_{\infty} a_n$? I am talking about the $\Pi_{\infty}a_{n}$ here. – user1952500 Aug 17 '17 at 01:19
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    By the way, we do not say products converge to zero, but rather, we say the diverge to zero. Zero of products are the $-\infty$ of sums. – Simply Beautiful Art Aug 17 '17 at 01:20
  • @user1952500 just the sequence. It doesn't matter if a sequence comes from a product or is by itself the conclusion is the same: just because a sequence is decreasing and bounded bellow by $c$ doesn't mean the sequence converges to $c$. – Trevor Gunn Aug 17 '17 at 01:27
  • @TrevorGunn I think the two are different and please correct me if I am wrong. In the case of your sequence, $\Pi_\infty a_n$ would not be a decreasing sequence. – user1952500 Aug 17 '17 at 01:29
  • @user1952500 $\prod_\infty a_n$ is not a sequence. – Trevor Gunn Aug 17 '17 at 01:30
  • @TrevorGunn, but $S_n = \Pi_{k=1}^n a_k$ is, and is a sequence bounded by $c$ and is continuously decreasing. Do you have an example of an $a_k$ where $c \ne 0$ ? – user1952500 Aug 17 '17 at 01:31
  • @user1952500 $a_n = (n + 1)/n$ is bounded below by $0$ and continuously decreasing but $a_n \to 1$ not $0$. – Trevor Gunn Aug 17 '17 at 01:33
  • @TrevorGunn I think I should ask a question to clarify the whole point. I am talking about the convergence / divergence of $S_n = \Pi_{k=1}^n a_n$ here. For the $a_n$ you mention, $S_n$ is continuously increasing. – user1952500 Aug 17 '17 at 01:35
  • @user1952500 Take $a_n = 1 - \frac{1}{n^2}$ then $S_n = \prod_{k=2}^n a_k \to \frac{1}{2}$. – Winther Aug 17 '17 at 01:35
  • @Winther thanks for that example! – user1952500 Aug 17 '17 at 01:36
  • @TrevorGunn thanks for pointing out the issue! – user1952500 Aug 17 '17 at 01:44

4 Answers4

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The given product equals $$\exp\sum_{n\geq 1}\log\frac{\log(2n)}{\log(2n+1)}=\exp\sum_{n\geq 1}-\int_{2n}^{2n+1}\frac{dx}{x\log x}\\=\exp\left[C-\sum_{n\geq 2}\frac{1}{(2n+\eta_n)\log(2n+\eta_n)}\right]$$ with $\eta_n\in(0,1)$ by the mean value theorem. Since both the series $\sum_{n\geq 2}\frac{1}{(2n+1)\log(2n+1)}$ and $\sum_{n\geq 2}\frac{1}{2n\log(2n)}$ are divergent by Cauchy's condensation test, the given product is convergent to zero, albeit very slowly.

Jack D'Aurizio
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Hint:

Since

$$\ln(1+x)\sim_0x$$

It follows from logarithmic rules that

$$\frac{\ln(2n+1)}{\ln(2n)}=1+\frac{\ln(1+1/2n)}{\ln(2n)}\sim_\infty1+\frac1{2n\ln(2n)}$$

Reciprocate the product, then take the log of it, and with this, we find

$$-\ln\prod_{k=1}^\infty\frac{\ln(2k)}{\ln(2k+1)}\sim\sum_{k=1}^\infty\frac1{2n\ln(2n)}=+\infty$$

Thus, the given product should go to

$$\prod_{k=1}^\infty\frac{\ln(2k)}{\ln(2k+1)}=e^{-\infty}=0$$

This approach may be rigorously done using squeezing and the bounds

$$x\ln(2)<\ln(1+x)<x\quad\forall x\in(0,1)$$

Simply Beautiful Art
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$$\ln \left(\dfrac{\ln(2n)}{\ln(2n+1)}\right) \sim \ln\left(1 - \frac{1}{2 n \ln(n)}\right) \sim - \frac{1}{2 n \ln(n)}$$ and $\sum_n \frac{1}{n \ln(n)}$ diverges, so the limit is $0$.

Robert Israel
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I just read this question and basically we can use the same approach.

If we denote $$P_n=\prod_{i=1}^n \frac{ln(2i)}{ln(2i+1)} \ and \ Q_n=\prod_{i=1}^n \frac{ln(2i+1)}{ln(2i+2)} \ then \ P_n*Q_n = \frac{ln(2)}{ln(2n+2)} \, now \ Q_n \ge P_n \ge 0 \\ thus \ 0 \le P_n \le \sqrt{\frac{ln(2)}{ln(2n+2)}}$$

motoras
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