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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Why do you say it is not true? –  Teake Nutma Jul 21 '13 at 12:09
@Pinguin Dirk Consider $\sqrt{a}$ is an integer. –  Chris's sis Jul 21 '13 at 12:12
@TeakeNutma see the message above. –  Chris's sis Jul 21 '13 at 12:13
Simpler example: In[4]:= Limit[(Sin[x]/x)^(1/x), x -> Infinity] Out[4]= 1 I'm not sure if this is a bug or a feature. Will investigate when time permits. –  Daniel Lichtblau Jul 21 '13 at 16:38
@DanielLichtblau Daniel, do note that when operating with real/rational numbers we end up with a 1 and when we add the assumptions about the integers we get the 0. That's regarding the OP's question. –  Sektor Jul 22 '13 at 9:08

$$\prod_{n=1}^\infty\left|\,1-\frac{a}{n^2}\,\right|^{1/\sqrt{a}}$$ 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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did you consider the case when $\sqrt{a}$ is an integer? –  Chris's sis Jul 21 '13 at 12:22
If you take any $\sqrt{a}$ integer then the limit is precisely $0$. $$\pi ^{-\frac{1}{\sqrt{a}}} \left(\frac{\sin \left(\pi \sqrt{a}\right)}{\sqrt{a}}\right)^{\frac{1}{\sqrt{a}}}$$ –  Chris's sis Jul 21 '13 at 12:29
@Chris'ssister: and the expression you give there goes to $1$ as $a\to\infty$ –  robjohn Jul 21 '13 at 12:31
Definitely no. It's precisely 0 when $\sqrt{a}$ is integer. –  Chris's sis Jul 21 '13 at 12:32
Oh, yes. one term is 0. So the limit doesn't exist. –  robjohn Jul 21 '13 at 12:32

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, Infinity}])^(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:

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, Infinity}])^(1/Sqrt[a]),
a -> Infinity, Assumptions :> Sqrt[a] \[Element] Integers]

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What if $\sqrt{a}$ is an integer? –  Chris's sis Jul 21 '13 at 12:20
@Chris'ssister How about now ? –  Sektor Jul 21 '13 at 12:24
Thanks so much! Why do you think Mathematica says the limit exists when it doesn't? –  Chris's sis Jul 21 '13 at 12:26
Technically speaking the limit still exists before we made the assumptions, but the "problem" is Mathematica was operating with Real numbers. –  Sektor Jul 21 '13 at 12:34
Thanks for your effort (+1) –  Chris's sis Jul 21 '13 at 12:36