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I was wondering whether the infinite product

$$\prod_{i=0}^{\infty}\frac{2n + 2}{2n + 1} = \frac{2}{1}\times\frac{4}{3}\times\frac{6}{5}\times\dots$$

converges. For all I know it is certainly not absolutely convergent, but is it at least conditionally so? Telescoping is not possible in this case. I know that

$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \ln{2}$$

but if I take the logarithm off this infinite product, I'd get

$$ \ln{1} - \ln{\frac{1}{2}} + \ln{\frac{1}{3}} - \ln{\frac{1}{4}} + \dots$$

which is not conclusive.

Voile
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    All the terms are greater than $1$, so there is no difference between conditional and absolute convergence. Maybe you had something else in mind? – Eric Wofsey Jul 16 '18 at 06:52
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    Not a duplicate, but here is another question that use the same expression: https://math.stackexchange.com/questions/855990/if-k-frac21-times-frac43-times-cdots-times-frac10099-then – user061703 Jul 16 '18 at 06:55
  • @EricWofsey: It doesn't answer the question: I can arrange it as $\ln{\frac{2}{1}} + ln{\frac{4}{3}} + ln{\frac{6}{5}} + \dots$ where the terms approachs 0, and I think ratio test gives 1. – Voile Jul 16 '18 at 06:55
  • http://ramanujan.math.trinity.edu/wtrench/research/papers/TRENCH_RP_93.PDF – saulspatz Jul 16 '18 at 06:58
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    Probably not much of a help, but it's basically $\lim_{n\to \infty} \frac{(2^n n!)^2}{(2n)!}$. – Sil Jul 16 '18 at 07:07
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    @Voile But it does answer the question... Since all terms are strictly greater than $1$, if it is conditionally convergent, it must also be absolutely convergent. Since you say that it is not absolutely convergent, it follows that it is not conditionally convergent. The product diverges. – Eff Jul 16 '18 at 07:11
  • Check out Wikipedia. – Jyrki Lahtonen Jul 16 '18 at 07:20
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    Of no use, but the product up to $p$ is $\approx \sqrt{p \pi}$ for large values of $p$. – Claude Leibovici Jul 16 '18 at 08:59

1 Answers1

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Hint: If $a,b,c,\ldots$ are all positive $$(1+a)(1+b)(1+c)\cdots\ge 1+a+b+c+\cdots.$$

Jyrki Lahtonen
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    The general theme at play here is to always compare the convergence of $\prod_i(1+a_i)$ and $\sum_ia_i$. Surprisingly I didn't find this covered explicitly in our question abouot infinite products. – Jyrki Lahtonen Jul 16 '18 at 07:15