10
$\begingroup$

$$\prod_{k=0}^{\infty} \biggl(1+ {\frac{1}{2^{2^k}}}\biggr)$$

My teacher gave me this question and said that this is easy only if it strikes the minute you read it. But I'm still thinking. Help!

P.S. This question is to be attempted by telescopic method.

$\endgroup$
  • $\begingroup$ I suggest multiply a few terms of the product and rearrange to see. $\endgroup$ – Maesumi Aug 9 '13 at 16:28
  • $\begingroup$ No luck @Maesumi. $\endgroup$ – Rohinb97 Aug 9 '13 at 16:29
  • 3
    $\begingroup$ Or multiply with $(1-1/2)$ and use $(A-B)(A+B)=A^2-B^2$ repeatedly. $\endgroup$ – Maesumi Aug 9 '13 at 16:31
9
$\begingroup$

The terms of the product are $(1+1/2)(1+1/4)(1+1/16)(1+1/256)\cdots$ with each denominator being the square of the previous denominator. Now if you multiply the product with $(1-1/2)$ you see telescoping action:

$(1-1/2)(1+1/2)=1-1/4$

$(1-1/4)(1+1/4)=1-1/16$

$(1-1/16)(1+1/16)=1-1/256$

Do you see the pattern developing?

$\endgroup$
  • 2
    $\begingroup$ How exactly did you think of this? $\endgroup$ – dfeuer Aug 9 '13 at 16:38
  • 1
    $\begingroup$ I multiplied the first few terms and saw that all $1/2^i$ show up in the product (that would be a straightforward way of answering the problem and perhaps more illuminating). So I sort of made my feet wet by actually trying to calculate whatever the product was. But to explain the process in way that looks like "telescoping" I thought of the difference of square identity. $\endgroup$ – Maesumi Aug 9 '13 at 17:07
7
$\begingroup$

$$1+ {\frac{1}{2^{2^k}}}=\cfrac{1- {\cfrac{1}{2^{2^{k+1}}}}}{1- {\cfrac{1}{2^{2^k}}}}=\frac{u_{k+1}}{u_k}$$ hence

$$\prod_{k=0}^{\infty} \biggl(1+ {\frac{1}{2^{2^k}}}\biggr)=\frac{1}{u_0}=2$$

$\endgroup$
  • $\begingroup$ Are you sure this is right? $\endgroup$ – Pratyush Sarkar Aug 9 '13 at 16:49
  • $\begingroup$ Yes I think so. $\endgroup$ – user63181 Aug 9 '13 at 16:50
  • $\begingroup$ Oh, now you fixed it. It was not right before. $\endgroup$ – Pratyush Sarkar Aug 9 '13 at 16:51
  • $\begingroup$ If you multiply the expressions on LHS and the expressions on RHS in the equalities given by Maesumi's answer and then cancelling we find $(1-1/2) (1+1/2)(1+1/4)(1+1/16)(1+1/256)\cdots=1$ so your product is $2$ $\endgroup$ – user63181 Aug 9 '13 at 17:11
  • $\begingroup$ I know, I was saying your answer wasn't correct because before your edit you had $\frac{8}{3}$ as the answer for some reason. $\endgroup$ – Pratyush Sarkar Aug 9 '13 at 17:25
2
$\begingroup$

$(1-1/2)(1+1/2)=1-1/2^2$

$(1-1/2^2)(1+1/2^2)=1-1/2^4$

and so on.

Then you get $(1-1/2)\prod_{k=0}^{n}(1+ 1/2^{2^k})=1- 1/2^{2^{k+1}}$.

$\endgroup$
  • $\begingroup$ The second line is wrong. $\endgroup$ – dfeuer Aug 9 '13 at 20:06
2
$\begingroup$

Here's what struck me the minute I read the problem (and before I read you were supposed to use a telescoping method):

$$\begin{align} \prod_{k=0}^\infty\left(1+{1\over2^{2^k}}\right) &= \left(1+{1\over2^1}\right)\left(1+{1\over2^2}\right)\left(1+{1\over2^4}\right)\cdots\\ &= 1+{1\over2^1}+{1\over2^2}+\cdots+{1\over2^{1+2}}+{1\over2^{1+4}}+\cdots+{1\over2^{1+2+4}}+{1\over2^{1+2+8}}+\cdots\\ &=1+{1\over2}+{1\over4}+{1\over8}+\cdots\\ &=2 \end{align}$$

using the fact that when you expand the product into the first sum, the exponents in the powers of $2$ are simply the positive integers written in base-$2$ form. This is a little bit like the way the unique factorization of numbers into primes is used to prove the product formula for the Riemann zeta function.

$\endgroup$

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.