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A product is defined for $n\in\mathbb{N},$ as

$$a_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdot...\cdot\left(1-\frac{1}{n^2}\right).$$

a) Show that the limit $\lim_{x\rightarrow \infty}a_n$ exists.

b) Compute the limit $\lim_{x\rightarrow \infty}a_n$.

a) For the limit to exist, we have to show that the decreasing sequence is bounded below. We have that $a_n>0$ and

$$\frac{a_{n+1}}{a_n}=\left(1-\frac{1}{n^2}\right)<1\Longleftrightarrow a_{n+1}=a_n\left(1-\frac{1}{n^2}\right)<a_n,$$

This shows that the limit exists.

Is this line of thought correct?

b) Here I'm stuck. I'm not sure what to do next. I tried to express $\ln(a_n)$ as a sum but to no avail.

NOTE: No expansions of any sort is to be used. Only elementary calculus.

Parseval
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    Hint: $1 - \frac{1}{n^2} = \frac{n^2-1}{n^2} = \frac{(n-1)(n+1)}{n^2}$. – fractal1729 Dec 21 '17 at 20:50
  • I did that rewriting too, but I could not see anyhing useful. – Parseval Dec 21 '17 at 20:53
  • for a) Im not sure if this line of thought is correct but you can use other convergence test for $\sum_k \ln(a_k)$. For b) the product seems related with the sine function if I remember correctly. – Masacroso Dec 21 '17 at 20:54
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    The point is it telescopes. $$ \frac{(2-1)(2+1)}{2^2} \frac{(3-1)(3+1)}{3^2} \ldots \frac{(n-1)(n+1)}{n^2} = \frac{n+1}{2n}$$ – Robert Israel Dec 21 '17 at 21:03
  • Writing it as $\displaystyle\frac{(n-1)/n}{n/(n+1)}$ would make it too obvious. BTW, one could also use the infinite product representation of $\displaystyle\frac{\sin x}x$. –  Dec 21 '17 at 21:05

4 Answers4

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\begin{eqnarray*} \frac{1 \times \color{red}{3}}{2 \times \color{blue}{2}} \frac{\color{blue}{2} \times \color{red}{4}}{\color{red}{3} \times \color{blue}{3}} \cdots \frac{\color{blue}{(n-2)} \times \color{red}{n}}{\color{red}{(n-1)} \times \color{blue}{(n-1)}} \frac{\color{blue}{(n-1)} \times \color{red}{(n+1)}}{\color{red}{n} \times \color{blue}{n}} \frac{\color{blue}{n} \times \color{red}{(n+2)}}{\color{red}{(n+1)} \times \color{blue}{(n+1)}} \cdots \end{eqnarray*}

Donald Splutterwit
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Maybe take a look at the function

$$f(x)=\left(1-\frac{x^2}{2^2}\right)\left(1-\frac{x^2}{3^2}\right)\cdots$$ or

$$f(x)=\cdots(1+x/3)(1+x/2)(1-x/2)(1-x/3)(1-x/4)\cdots$$ which has periodic zeroes at all integers except at zero, 1 and −1. We know that analytical functions match (up to a constant factor) if they match in all their poles and zeroes, so we know which function this is. Periodic zeroes suggest that it's a trigonometric function, divided out by some factors to remove 3 zeroes, and value $f(0)$ tells you the pre-factor. Then just calculate $f(1)$.

$\sin(\pi x)$ has zeroes at $x=\cdots -3,-2,-1,0,1,2,3\cdots$. $\sin(\pi x)/(x(1-x^2))$ has zeroes at $\cdots -3,-2,2,3,\cdots$, but has a value $\pi$ at $x=0$. Try:

$$f(x)=\frac{\sin \pi x}{\pi x (1-x^2)}$$

and compute

$$\lim_{x\to 1}f(x)$$

With less calculus, just put everything on the same denominator:

$$a_n=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)\cdots(n-1)(n+1)}{(n!)^2}$$

$$a_n=\frac{1\cdot 3\cdot 2\cdot 4\cdot 3\cdot 5\cdots (n-1)(n+1)}{(n!)^2}$$

$$a_n=\frac{1\cdot 2\cdot (3\cdot 4\cdot 5\cdots (n-1))^2 \cdot n(n+1)}{(n!)^2}$$ $$a_n=\frac{(n!)^2/(2 n)\cdot (n+1)}{(n!)^2}=\frac{n+1}{2n}\to \frac12$$

orion
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HINT : what about $\frac{(n-2)n}{(n-1)^{2}}\frac{(n-1)(n+1)}{n^{2}}\frac{n(n+2)}{(n+1)^{2}}$

openspace
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There is something quite hand-wavy you can do by refactoring it into ${\prod_2^\inf\frac{(n+1)(n-1)}{n^2}}$.

Write out the first few products in the sequence:

${\frac{3*1}{2^2}*\frac{4*2}{3^2}*\frac{5*3}{4^2}...}$

Notice that for the first term, the 2 in the denominator is cancelled by the 2 in the numerator of the second. Leaving just a single 2 in the denominator.

Now turn your attention to the ${3^2}$ in the denominator of the second term. It's cancelled (divides to 1) with the 3 in the numerator of the first and the 3 in the numerator of the 3rd. Same with the 4 in the denominator in the 3rd term, and etc...

${\frac{\not3*1}{2^{\not2}}*\frac{\not4*\not2}{\not3^{\not2}}*\frac{\not5*\not3}{\not4^{\not2}}*\frac{6*\not4}{5^{\not2}}...}$

What's left after all these things cancel (become 1)? Just the 2 in the denominator of the first term. ${\prod_2^\inf(1-\frac{1}{n^2})\rightarrow\frac{1}{2}}$

James Bender
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  • damn... I should learn to write things out like that... makes sense now. Thanks! – Parseval Dec 21 '17 at 21:37
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    Actually, you should be very careful writing things out in this way. It's more a technique to guide your intuition about what the answer may be. Modern analysis has shown 10x over when this kind of reasoning can and cannot be used. Even in the cases where it can be used, don't take it as a proof without adding in the theorems that show this particular case to be valid. – James Bender Dec 21 '17 at 22:13
  • What I meant was that it never hurts to try this method :) I lost 3 points on my exam because I did not do this. – Parseval Dec 22 '17 at 13:16