Simplifying to a Geometric Series

[xago]

Homework Statement

I have a question with asks to solve a differential equation via power series and I've done everything up to finding the recurrence relation which is a_{n+2} = -\frac{a_{n}}{n+2}
Given the initial conditions a_{o} = 1 and a_{1} = 0 I'm trying to simplify the series into a geometric series. The series is 1,-1/2, 1/8, -1/48, 1/480, -1/5760 etc...

The Attempt at a Solution

So far I've gotten (-1)^{n} in the numerator to account for the alternating negative sign, however I can't find the denominator for the life of me...
my best attempt is \frac{(-1)^{n}}{(2^n)} but it only works for the first 2 terms :S

 

[Mute]

Don't just try to randomly guess what the pattern is by looking at the first few terms. You want to try to do it a little more systematically than that. Try to write a_{n+2} in terms of a_0 (hopefully you noticed that all the odd coefficients are zero).

i.e., start writing
a_{2k+2} = -\frac{a_{2k}}{2k+2} = (-1)^2\frac{a_{2k-2}}{(2k+2)(2k)} = (-1)^3\frac{a_{2k-4}}{(2k+2)(2k)(2k-2)} = \dots
and then see if you can guess a pattern that gives a_{2k+2} = a_0/(\mbox{stuff}). (note that I set n = 2k).

Once you figure out the pattern, if you want to legitimately prove it's correct you can prove it using induction (i.e., prove it holds for k = 1, then prove that if you assume the form is true for k it follows that it's true for k+1).

Also, note that the terms in a geometric series have the form ar^k. Your terms won't look quite like this, so it's not exactly a geometric series.

 

[xago]

The best I can do is 2k! which gives 2k(2k-2)(2k-4)...right? it works up k=3 to but it's missing the 2k+2 term and doesn't work for k=4
I know 2k! is expressed as 2^{k}k! but how is (2k+2)! expressed in such a way that I can calculate it?

edit*
oops, 2k! is right I believe. I miscalculated the series in my first post, instead of -1/48, 1/480 it should have been -1/48, 1/384 etc..

 

[obafgkmrns]

Looks like 1/480 should be 1/384 instead, and 1/5760 be 1/3840. Did you skip a term?

In any case, you might want to check out Sloan's Sequences: https://oeis.org/

 

[Mute]

The best I can do is 2k! which gives 2k(2k-2)(2k-4)...right? it works up k=3 to but it's missing the 2k+2 term and doesn't work for k=4
I know 2k! is expressed as 2^{k}k! but how is (2k+2)! expressed in such a way that I can calculate it?

If you're still not sure, just take (2k)! and replace k with k + 1: (2(k+1))! = (2k+2)! = 2^k+1(k+1)!, such that

a_{2k+2} = \frac{(-1)^{k+1}}{2^{k+1}(k+1)!}

 

[HallsofIvy]

The best I can do is 2k! which gives 2k(2k-2)(2k-4)...right? it works up k=3 to but it's missing the 2k+2 term and doesn't work for k=4
I know 2k! is expressed as 2^{k}k! but how is (2k+2)! expressed in such a way that I can calculate it?

(2k+2)!= (2(k+1))! so it would just be 2^{k+1}(k+1)!

edit*
oops, 2k! is right I believe. I miscalculated the series in my first post, instead of -1/48, 1/480 it should have been -1/48, 1/384 etc..