Factorial Series... please help me with my calculus!?

Question

I understand simple factorials, but not factorial quotients where you have a numerator and denominator that don't simplify. I don't know what to do with expressions like 1 * 3 * 5 * 7 ..... (2n - 1).

There are two alternating series I'm stuck on. I am supposed to determine the convergence or divergence of the series using the Alternating Series Test: find where an is > 0; the limit of an as n approaches infinity must = 0; and you must show that a(n + 1) [the next nth term after an] is less than an for all values of n. We've been using derivatives to show a function is decreasing so I don't understand these mathematical induction style proofs.

1) Summation, from 1 to infinity:
[(-1)^(n + 1) * n!]/[1 * 3 * 5 . . . (2n - 1)]

2) Summation, from 1 to infinity :
[(-1)^(n + 1) * 1 * 3 * 5 . . . (2n - 1)]/[1 * 4 * 7 * . . . (3n - 2)]

If any math genius happens to see this, I would really be happy to have some help. Thanks. :)

Answer 1

Taking points in order...

1*3*5*7*...*(2n-1) is often written (2n-1)!!, the "double factorial" of (2n-1). You can express it in terms of normal factorials as (2n-1)! / [2^(n-1).(n-1)!], or more compactly as (2n)! / (2^n . n!). The double factorial of an even number is simpler: (2n)!! = 2^n . n!. Of course (2n)!! . (2n-1)!! = (2n)!.

Anyway, all that is pretty much irrelevant to the alternating series thing.
1) (-1)^(n+1) a_n, where a_n = n! / (2n-1)!!
a_(n+1) = (n+1)! / (2n+1)!!
So a_(n+1) / a_n = (n+1) / (2n+1) < 1, so a_(n+1) < a_n for all n.
Also, lim (n->∞) a_(n+1) / a_n
= lim (n->∞) (n+1) / (2n+1)
= 1/2
Since this is strictly less than 1, lim (n->∞) a_n = 0.

2) is fairly similar; the ratio of the (n+1)'th term to the n'th term is (2n+1) / (3n+1) which is less than 1 and tends to 2/3 as n->∞, so the terms are decreasing and tend to 0.

Answer 2

for number 1
I find the ratio test to be helpful in these instances.

|((n+1)!/(1*3*5*(2(n+1)-1)))/(n!/(1*3*5*(2n+1))|

using the ratio test we can get rid of the (-1)^(n+1) because of the absolute value

now just so ya know (n+1)!/n! = n+1
say n = 5 you would have (6*5*4*3*2)/(5*4*3*2) = 6. makes sence right?

so now using the same logic ew do that to the long multiplications.

(1*3*5*(2n+1))/(1*3*5*(2(n+1)-1)) = 1/(2(n+1)-1) = 1/(2n+1)

This bit is the same logic as in the factorial so now we combine the two pieces above. This whole process is just algebra.

The way i set them up we can multiply the two together to get 6/(2n+1)

the ratio test asks for the limit as n-> infinity of 6/(2n+1) which is 0 which is less than 1
Therefore number 1 converges



Now on to number two

I'm going to skip some steps but it should make sence.

using the ratio test and canceling out all the terms we will get

limit as n -> infinity of (2n+1)/((3n+1) = 2/3 < 1
Therefore it too convcerges

If you understand the concept of (n+a)!/n! = n+a where a is an arbitrary constant, you will be able to reduce these recursive multiplicating functions easily.

you can also turn the formula's into a recursive one and solve using the definition of the ratio test
the following is an example


a\/1 = 2 (i use \/ to denote a subscript)

a\/(n+1) = (2+ cos(n)/n) * a\/n

using the definition

limit as n -> infinity of | a\/(n+1) / a\/n |
= said limit of ((2+ cos(n)/n) * a\/n) / a\/n |
= said limit of (2+ cos(n)/n)
= 0 < 1
Converges

Answer 3

I think denominator is (2k)! k=0 0! =1 k =1 2! k=2 4! k=3 6! for all the four terms it works out ok