# General term of the sequence, if it exists

1. Mar 13, 2005

### relinquished™

Hello. I've been having such a hard time thinking of the general term of this "Sequence". Actually, I'm not even sure if this is a sequence at all, but it looks like it can be simplified into one summation symbol.
$$\frac{-2}{6}, \frac{-20}{120}, \frac{-1080}{5040}, \frac{-140400}{362880}, ...$$

The denominators of every term are actually the factorials of the odd numbers starting from 3, what i can't find is the "pattern" for the numerator.

Thanks for any help :)

2. Mar 14, 2005

### damoclark

There's a nice data base of integer sequences at :

http://www.research.att.com/~njas/sequences/ [Broken]

that you can search.

I played around with the sequence you have given, but couldn't figure anything much out. Do you know anymore of the terms???

Last edited by a moderator: May 1, 2017
3. Mar 14, 2005

### relinquished™

Actually, with a little tinkering I did manage to find the pattern :) I just have one question... It is related to the sequence but its not actually the sequence

is this statement true?

$$\prod_{n=0} (2n+1) = (1)(3)(5)(7)(9)...$$

Note: The n in the "Prod" symbol tends to infinity. I dont know how to place an upper limit in the symbol XD

I'm not so familiar with the symbol, i just saw it in the HowToLaTeX FAQ and wondered if it's like the summation symbol (only it means product) :)

Thanks again for that site. It did help me in a way :)

4. Mar 14, 2005

### Muzza

I guess the statement is true, but it doesn't appear to be well-defined...

If you want to know how to place an upper limit:

$$\prod_{n=0}^{\infty} (2n+1)$$

5. Mar 14, 2005

### relinquished™

so it should be

$$\prod_{n=0}}^{\infty} (2n+1) = (1)(3)(5)(7)(9)(11)(13).....$$

Is it "more" well defined now?

Thanks again

6. Mar 14, 2005

### Muzza

No, it just seems like it's "meaningless" to talk about the product of all odd natural numbers ;)

Last edited: Mar 14, 2005
7. Mar 14, 2005

### relinquished™

well, yeah, it is meaningless. But when it becomes a part of a general term of a series that is a solution to a differential equation it is kinda important :)

which leads me to my last question, (which I know should be part of Differential Equations but my main focus was simplifying the general term of a series) in most differential equations books when I read their solutions they write their general term as (1)(3)(5)(7)...(2n+1) (If the need or occasion arose). My question is if it's more appropriate to write it as

$$\prod_{n=0}^{\infty} (2n+1)$$

Thanks a bunch :)

8. Mar 14, 2005

### shmoe

If your general term is (1)(3)(5)...(2n+1) then you could write it as

$$\prod_{i=0}^{n}(2i+1)$$

Note the endpoints carefully. Either one is fine as long as there's no ambiguity for what the ... represent. My preference is towards the $\prod$ notation as long as there are no typsetting issues.

9. Mar 14, 2005

### relinquished™

Thanks for everything :)

10. Mar 14, 2005

### Data

An alternative notation, that is sometimes prettier, and that doesn't involve product notation:

$$\prod_{i=0}^n (2i+1) = \frac{(2n+1)!}{2^n n!}$$

Edit: Actually, looking at your situation, this notation might lead to some simplifications too!

Last edited: Mar 14, 2005