# Thread: Value of infinite products

1. ## Value of infinite products

Hi, I've been asked to find the value of some infinite products. We've done infinite series before, but not products. There's a few questions - if someone could help me out with the first one, I could try the other two myself.

Question: Find the values of the following infinite products:

Cheers!

2. Originally Posted by mathsperson
Hi, I've been asked to find the value of some infinite products. We've done infinite series before, but not products. There's a few questions - if someone could help me out with the first one, I could try the other two myself.

Question: Find the values of the following infinite products:

Cheers!
Your first two can be found here Infinite Product -- from Wolfram MathWorld

and your third can be written in closed form

$\frac{1}{1-x^4}$

3. Thanks. What method would be used to evaluate these? I've done infinite sums before, but these are the first questions I've seen on infinite products.

4. Originally Posted by mathsperson
Thanks. What method would be used to evaluate these? I've done infinite sums before, but these are the first questions I've seen on infinite products.
You know how you can sometimes sum a series by forming a "telescoping sum", where you manipulate the terms (maybe by using partial fractions) so that each term cancels with those on either side of it? Well, to evaluate an infinite product, you can sometimes form a "telescoping product", where something similar happens.

For example, $1 - \frac1{n^2} = \frac{n^2-1}{n^2} = \frac{n-1}n\,\frac{n+1}n$. Therefore

$\prod_{n=2}^N\left(1-\frac1{n^2}\right) =$ $\left(\frac12\frac{\rlap{\color{red}/}3}{\rlap{\color{blue}/}2}\right) \left(\frac{\rlap{\color{blue}/}2}{\rlap{\color{red}/}3}\frac{\rlap{\color{red}/}4}{\rlap{\color{blue}/}3}\right) \left(\frac{\rlap{\color{blue}/}3}{\rlap{\color{red}/}4}\frac{\rlap{\color{red}/}5}{\rlap{\color{blue}/}4}\right) \cdots \left(\frac{\rlap{\color{blue}\vrule height3pt depth-2.5pt width2.3em}N-1}{\rlap{\color{red}\,/}N}\frac{N+1}N\right)$ $= \frac12\frac{N+1}N \to\frac12$ as $N\to\infty$.

For $\prod_{n=2}^\infty\frac{n^3-1}{n^3+1}$ you can do the same thing. Factorise $\frac{n^3-1}{n^3+1}$ as $\frac{n-1}{n+1}\,\frac{n^2+n+1}{n^2-n+1}$, write down the product of the first few terms and see how they telescope. You should then be able to show that the product of the first N terms converges to 2/3 as N→∞.

5. Originally Posted by mathsperson
Hi, I've been asked to find the value of some infinite products. We've done infinite series before, but not products. There's a few questions - if someone could help me out with the first one, I could try the other two myself.

Question: Find the values of the following infinite products:

Cheers!
Also you could do the following

\begin{aligned}\prod_{n=2}^{\infty}\left(1-\frac{1}{n^2}\right)&=\exp\left(\ln\prod_{n=2}^{\i nfty}\left(1-\frac{1}{n^2}\right)\right)\\
&=\exp\left(\sum_{n=2}^{\infty}\ln\left(1-\frac{1}{n^2}\right)\right)\\
&=\exp\left(-\ln(2)\right){\color{red}\star}\\
&=\frac{1}{2}\end{aligned}

${\color{red}\star}\sum \ln\left(1-\frac{1}{n^2}\right)=\sum\left\{\ln\left(1+\frac{1 }{n}\right)+\ln\left(1-\frac{1}{n}\right)\right\}\overbrace{\longmapsto}^ {\text{telescopes}}-\ln(2)$

Its a neat alternative to Opalg's solution...although admittedly not as clean

6. Another solution

Let $p_n=\prod_{k=2}^{n}\left\{1-\frac{1}{k^2}\right\}$

Proposition: The n-th partial product is $p_n=\frac{n+1}{2n}$

Proof:

Base case: $p_2=\frac{3}{4}=\frac{2+1}{2\cdot2}$

Inductive Hypothesis: $p_n=\frac{n+1}{2n}$

Inductive step:

\begin{aligned}p_{n+1}&=p_n\cdot \left(1-\frac{1}{(n+1)^2}\right)\\
&=\frac{n+1}{2n}\cdot\frac{(n+1)^2-1}{(n+1)^2}\\
&=\frac{n^2+2n}{2n(n+1)}\\

Therefore

\begin{aligned}\prod_{k=2}^{\infty}\left\{1-\frac{1}{k^2}\right\}&=\lim_{n\to\infty}p_n\\
&=\lim_{n\to\infty}\frac{n+1}{2n}\\
&=\frac{1}{2}\end{aligned}

7. For the second one (as Mathstud did), if we let

$
p_n = \prod_{k=2}^{n} \frac{k^3-1}{k^3 + 1}$

then by induction we can show

$
p_n = \frac{2}{3}\, \frac{n^2+n+1}{n(n+1)}$

as the rest follows.

8. Originally Posted by danny arrigo
Your first two can be found here Infinite Product -- from Wolfram MathWorld

and your third can be written in closed form

$\frac{1}{1-x^4}$
More details on the last one and it's answer. Consider

$(1-x^4)(1+x^4) = 1 - x^8$

$\underbrace{(1-x^4)(1+x^4)}_{\text{from above} \, = \,(1-x^8)}(1+x^8) = (1-x^8)(1+x^8) = 1 - x^{16}$

$\underbrace{(1-x^4)(1+x^4)(1+x^8)}_{\text{from above} \, = \,(1-x^{16})}(1+x^{16}) = (1-x^{16})(1 + x^{16}) = 1- x^{32}$

which generalizes (and can be proven by induction)

$(1-x^4)(1+x^4)(1+x^8) \cdots (1+x^{2N}) = 1 - x^{2(N+1)}$.

Hence,

$(1+x^4)(1+x^8) \cdots (1+x^{2N}) = \frac{1- x^{2(N+1)}}{1-x^4}$,

or

$\prod_{n=2}^N 1+x^{2n} = \frac{1- x^{2(N+1)}}{1-x^4}$.

If $|\,x\,| <1$, then in the limit as $N\, \rightarrow\, \infty$

$\prod_{n=2}^{\infty} 1+x^{2n} = \frac{1}{1-x^4}$.

(Sorry, I just couldn't leave it without explanation)

9. Originally Posted by danny arrigo
More details on the last one and it's answer. Consider

$(1-x^4)(1+x^4) = 1 - x^8$

$\underbrace{(1-x^4)(1+x^4)}_{\text{from above} \, = \,(1-x^8)}(1+x^8) = (1-x^8)(1+x^8) = 1 - x^{16}$

$\underbrace{(1-x^4)(1+x^4)(1+x^8)}_{\text{from above} \, = \,(1-x^{16})}(1+x^{16}) = (1-x^{16})(1 + x^{16}) = 1- x^{32}$

which generalizes (and can be proven by induction)

$(1-x^4)(1+x^4)(1+x^8) \cdots (1+x^{2N}) = 1 - x^{2(N+1)}$.

Hence,

$(1+x^4)(1+x^8) \cdots (1+x^{2N}) = \frac{1- x^{2(N+1)}}{1-x^4}$,

or

$\prod_{n=2}^N 1+x^{2n} = \frac{1- x^{2(N+1)}}{1-x^4}$.

If $|\,x\,| <1$, then in the limit as $N\, \rightarrow\, \infty$

$\prod_{n=2}^{\infty} 1+x^{2n} = \frac{1}{1-x^4}$.

(Sorry, I just couldn't leave it without explanation)
Are you sure you did not prove,
$\prod_{n=2}^{\infty} (1+x^{2^n}) = \frac{1}{1-x^4}$

10. Originally Posted by ThePerfectHacker
Are you sure you did not prove,
$\prod_{n=2}^{\infty} (1+x^{2^n}) = \frac{1}{1-x^4}$
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