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I have a curiosity as regards the infinite product below. I wonder why Mathematica v.8.0. says
the limit is $1$. This is not true.

Limit[(Product[(1 - a/n^2), {n, 1, Infinity}])^(1/Sqrt[a]),  a -> Infinity]
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  • $\begingroup$ Why do you say it is not true? $\endgroup$ Jul 21, 2013 at 12:09
  • $\begingroup$ @Pinguin Dirk Consider $\sqrt{a}$ is an integer. $\endgroup$ Jul 21, 2013 at 12:12
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    $\begingroup$ V10.1 now gives the result Interval[{0, 1}]. $\endgroup$
    – Michael E2
    May 4, 2015 at 2:18
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    $\begingroup$ @MichaelE2 V12.x give 1 again. (Arguably better/worse result than Interval[{0,1}]?) $\endgroup$
    – Silvia
    Nov 28, 2019 at 14:58
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    $\begingroup$ @Silvia Probably worse in that mathematically the limit Limit[p, a -> Infinity], where p = (Sin[Sqrt[a] Pi]/(Sqrt[a] Pi))^(1/Sqrt[a]) is the product, does not exist unless one restricts the domain to Sin[Sqrt[a] Pi] != 0. OTOH, it might be considered better in that one could argue that the product "diverges" to zero when $\sqrt{a}$ is an integer, and that the domain of a should be so restricted. However, the limit is computed as 1 without such a restriction, and that is incorrect. $\endgroup$
    – Michael E2
    Nov 28, 2019 at 15:53

2 Answers 2

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$$ \prod_{n=1}^\infty\left|\,1-\frac{a}{n^2}\,\right|^{1/\sqrt{a}} $$ As $a\to\infty$, each term is like $1+\frac{\log(a)}{n^2\sqrt{a}}$ and as $\frac{\log(a)}{n^2\sqrt{a}}$ is absolutely summable to something around $\frac{\log(a)}{\sqrt{a}}\frac{\pi^2}{6}\to0$ I would say that the product limits to $1$, unless $a=n^2$ for some $n$. So the limit doesn't exist.

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  • $\begingroup$ did you consider the case when $\sqrt{a}$ is an integer? $\endgroup$ Jul 21, 2013 at 12:22
  • $\begingroup$ If you take any $\sqrt{a}$ integer then the limit is precisely $0$. $$\pi ^{-\frac{1}{\sqrt{a}}} \left(\frac{\sin \left(\pi \sqrt{a}\right)}{\sqrt{a}}\right)^{\frac{1}{\sqrt{a}}}$$ $\endgroup$ Jul 21, 2013 at 12:29
  • $\begingroup$ @Chris'ssister: and the expression you give there goes to $1$ as $a\to\infty$ $\endgroup$
    – robjohn
    Jul 21, 2013 at 12:31
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    $\begingroup$ Oh, yes. one term is 0. So the limit doesn't exist. $\endgroup$
    – robjohn
    Jul 21, 2013 at 12:32
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    $\begingroup$ It's nice to see you over here robjohn $\endgroup$
    – Mr.Wizard
    Jul 21, 2013 at 12:51
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Any ideas why it is not true ? Typo/s ?

Running Win 8, Mathematica 9.0.1 the result is the same and also on my Ubuntu desktop with Mathematica 9.0.0 and WolframAlpha: Check here

Computing the (Product[(1 - a/n^2), {n, 1, ∞}])^(1/Sqrt[a]) yields the result:

$$ \pi ^{-\frac{1}{\sqrt{a}}} \left(\frac{\sin \left(\pi \sqrt{a}\right)}{\sqrt{a}}\right)^{\frac{1}{\sqrt{a}}} $$

Taking the limit of it as $a \to \infty$ is equal to $1$

And plotting it:

Plot

Edit: Assuming $\sqrt{a}\in \mathbb{Z}$ and taking the limit of the product we end up with $0$:

Limit[(Product[(1 - a/n^2), {n, 1, ∞}])^(1/Sqrt[a]), 
      a -> ∞, Assumptions :> Sqrt[a] ∈ Integers]
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  • $\begingroup$ What if $\sqrt{a}$ is an integer? $\endgroup$ Jul 21, 2013 at 12:20
  • $\begingroup$ @Chris'ssister How about now ? $\endgroup$
    – Sektor
    Jul 21, 2013 at 12:24
  • $\begingroup$ Thanks so much! Why do you think Mathematica says the limit exists when it doesn't? $\endgroup$ Jul 21, 2013 at 12:26
  • $\begingroup$ Technically speaking the limit still exists before we made the assumptions, but the "problem" is Mathematica was operating with Real numbers. $\endgroup$
    – Sektor
    Jul 21, 2013 at 12:34
  • $\begingroup$ Glad to help :) $\endgroup$
    – Sektor
    Jul 21, 2013 at 12:37

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