limit of a series

[safak]

Hi all!

 

how can I approximate the series:

 

[math] S_{n} = \sum_{k=0}^{n}\binom{3n}{k} [/math] ?

 

The approximation is necessary, since no closed form can be found.

 

Solutions are:

 

[math] S_{n} \sim 2\binom{3n}{n} [/math]

 

or more precise

 

[math] S_{n} = \binom{3n}{n}(2-\frac{4}{n}+\mathcal{O}(\frac{1}{n^2})) [/math]

 

but how did I get there?

 

 

 

Thanks!

 

 

safak

[the tree]

This looks an awful lot like homework to me. You'll have to demonstrate that you've made an effort, before anyone else does.

[safak]

Well, it is no homework. I stumbled upon this in the book 'Concrete Mathematics' from Graham,Knuth,Patashnik page 439 about Asymptotics and I don't get it, so I considered asking someone. It just says 'here is a series which has no closed form but which can be approximated to (1) and (2).", (1) and (2) refering to the above answers. I totaly don't know how to calculate series.

 

This is a science forum - why do you even bother about someone asking a question she/he apperently can't solve themselves. Some people learn more if their questions are answered, their grades are given by their teachers, and it is no use banging my head against this series if I don't know how to solve, i will look no good in ten years except some divinity beams the answer into my head Since my main to be learned chapter is not solving series but Asymptotics.

[Bignose]

This is a science forum - why do you even bother about someone asking a question she/he apperently can't solve themselves. Some people learn more if their questions are answered, their grades are given by their teachers, and it is no use banging my head against this series if I don't know how to solve, i will look no good in ten years except some divinity beams the answer into my head Since my main to be learned chapter is not solving series but Asymptotics.

 

Because science is about learning, and a goal of this forum is to help people learn for themselves, not spoon feeding answers. The question does have a homework feel to it. This forum is explicit in that it won't provide homework answers directly. It goes against much of the goals of this forum.

 

To answer your question, I don't think that there is a way to calculate the values given. They are approximations, which means that they won't be exactly right, but may be close enough for some applications. Such as applications where you only need be within 1 or 2 orders of magnitude. You may be able to see the logic behind the approximations if you spread it out, but per the text you quoted, they aren't the exact answers to the series. There are approximations.