# Product of series

1. Aug 18, 2011

### mishrashubham

1. The problem statement, all variables and given/known data

Prove that
$$\prod_{n=0}^n 1+3^{-2^{n}}= \frac{3}{2}(1-3^{-2^{n+1}})$$

2. Relevant equations

3. The attempt at a solution

I am an idiot in math and I couldn't figure out where to start. I thought about using log to convert it into a summation series (I find sums easier to handle than products), but I couldn't progress. I just need a direction, a place to start from. I can hopefully take over from there.

Thanks

Last edited: Aug 19, 2011
2. Aug 19, 2011

### GreenPrint

What course is this for? Just out of curiosity, I have no idea how to answer the question, and am just wondering what course this is for. I wasn't introduced to basic series stuff tell Calculus II.

3. Aug 19, 2011

### mishrashubham

I don't know about the US equivalent, but I have this in Algebra, under "Sequences, Series and Progressions".

4. Aug 19, 2011

### NeroKid

just use the induction it is quite easy

5. Aug 19, 2011

### dynamicsolo

With infinite series or products, when in doubt, try writing out the first few terms. Set n = 1 , write the two terms in the product and multiply them out; what do you get? Now try it with n = 2 and multiply the third term times your result from n = 1. What sort of sum are you getting? Do you know an expression for the sum of a finite set of such terms?

It's true that induction would suffice to prove the relationship, which is all that is asked here. However, it would be nice to be able to establish the relationship, which in fact also turns out to be easy if you start multiplying out the terms of the product and looking at what that gives you...

6. Aug 19, 2011

### mishrashubham

It really was extremely easy using induction. Thanks for the idea. I should have thought about using induction earlier.

I always do that if I encounter series; to see if I get some pattern. But I couldn't figure it out in this question.

Thanks for the replies people.

7. Aug 19, 2011

### GreenPrint

What's the name of the giant N character and which n is placed on top of and n=0 is placed on the bottom of? Is this just a different way to write sigma, and is the same way of indicating a series from n=0 to n?

8. Aug 19, 2011

### Galron

Pretty much yes but a series of products instead of sums.

It's the symbol for a Number Product a giant pi symbol in fact, but then that is trivial.

Product of n=0...n ...

ETA: for clarity.

oops.

9. Aug 19, 2011

### GreenPrint

My bad that's for product series, sigma is for summation series... hmm interesting

10. Aug 19, 2011

### uart

It's a product symbol but it's not quite written correctly. It should be:

$$\prod_{k=0}^n \left( 1+3^{-2^{k}}\right) = \frac{3}{2}(1-3^{-2^{n+1}})$$

11. Aug 19, 2011

### Ray Vickson

I get the answer $$1 + 3^{-2^n}$$ because your expression equals $$(\prod_{n=0}^n 1) + 3^{-2^n}.$$ However, if you meant $$\prod_{k=0}^n (1 + 3^{-2^k}),$$ then that would give a different answer.

RGV

12. Aug 19, 2011

### mishrashubham

My apologies.

One more thing. How would I go about solving it had I not known the right hand side?

13. Aug 19, 2011

### dynamicsolo

If you write the first two terms of the product, you find

$$( 1 + 3^{-2^{0}} ) ( 1 + 3^{-2^{1}} ) = ( 1 + 3^{-1} ) ( 1 + 3^{-2} ) = ( 1 + 3^{-1} + 3^{-2} + 3^{-3} ) .$$

The first three terms yield

$$( 1 + 3^{-1} + 3^{-2} + 3^{-3} ) ( 1 + 3^{-2^{2}} ) = ( 1 + 3^{-1} + 3^{-2} + 3^{-3} ) ( 1 + 3^{-4} ) = ( 1 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} + 3^{-5} + 3^{-6} + 3^{-7}) .$$

What sort of finite series is this? How many terms are there is each successive multiplication? Is there a formula for the sum of the general finite series?