9
$\begingroup$

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]
|improve this question
$\endgroup$
  • $\begingroup$ Why do you say it is not true? $\endgroup$ – Teake Nutma Jul 21 '13 at 12:09
  • $\begingroup$ @Pinguin Dirk Consider $\sqrt{a}$ is an integer. $\endgroup$ – user 1357113 Jul 21 '13 at 12:12
  • 2
    $\begingroup$ V10.1 now gives the result Interval[{0, 1}]. $\endgroup$ – Michael E2 May 4 '15 at 2:18
  • 1
    $\begingroup$ @MichaelE2 V12.x give 1 again. (Arguably better/worse result than Interval[{0,1}]?) $\endgroup$ – Silvia Nov 28 at 14:58
  • 1
    $\begingroup$ @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 at 15:53
10
$\begingroup$

$$ \prod_{n=1}^\infty\left|\,1-\frac{a}{n^2}\,\right|^{1/\sqrt{a}} $$ 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.

|improve this answer
$\endgroup$
  • $\begingroup$ did you consider the case when $\sqrt{a}$ is an integer? $\endgroup$ – user 1357113 Jul 21 '13 at 12:22
  • $\begingroup$ If you take any $\sqrt{a}$ integer then the limit is precisely $0$. $$\pi ^{-\frac{1}{\sqrt{a}}} \left(\frac{\sin \left(\pi \sqrt{a}\right)}{\sqrt{a}}\right)^{\frac{1}{\sqrt{a}}}$$ $\endgroup$ – user 1357113 Jul 21 '13 at 12:29
  • $\begingroup$ @Chris'ssister: and the expression you give there goes to $1$ as $a\to\infty$ $\endgroup$ – robjohn Jul 21 '13 at 12:31
  • 1
    $\begingroup$ Oh, yes. one term is 0. So the limit doesn't exist. $\endgroup$ – robjohn Jul 21 '13 at 12:32
  • 1
    $\begingroup$ It's nice to see you over here robjohn $\endgroup$ – Mr.Wizard♦ Jul 21 '13 at 12:51
10
$\begingroup$

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]
|improve this answer
$\endgroup$
  • $\begingroup$ What if $\sqrt{a}$ is an integer? $\endgroup$ – user 1357113 Jul 21 '13 at 12:20
  • $\begingroup$ @Chris'ssister How about now ? $\endgroup$ – Sektor Jul 21 '13 at 12:24
  • $\begingroup$ Thanks so much! Why do you think Mathematica says the limit exists when it doesn't? $\endgroup$ – user 1357113 Jul 21 '13 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 '13 at 12:34
  • $\begingroup$ Glad to help :) $\endgroup$ – Sektor Jul 21 '13 at 12:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.