What is the product of this by telescopic method? $$\prod_{k=0}^{\infty} \biggl(1+ {\frac{1}{2^{2^k}}}\biggr)$$
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.
 A: 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?
A: $$1+ {\frac{1}{2^{2^k}}}=\cfrac{1- {\cfrac{1}{2^{2^{k+1}}}}}{1- {\cfrac{1}{2^{2^k}}}}=\frac{u_{k+1}}{u_k}$$
hence
$$\prod_{k=0}^{\infty} \biggl(1+ {\frac{1}{2^{2^k}}}\biggr)=\frac{1}{u_0}=2$$
A: $(1-1/2)(1+1/2)=1-1/2^2$
$(1-1/2^2)(1+1/2^2)=1-1/2^4$
and so on.
Then you get $(1-1/2)\prod_{k=0}^{n}(1+ 1/2^{2^k})=1- 1/2^{2^{k+1}}$.
A: 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.
