+47 Consider the infinite series defined by
∞∑n=1 2n!/2^2n
(a) What is the value of r from the ratio test?
(b) What does this r value tell you about the series?
I know that the series diverges but I am confused on how to find the value of r from a ratio test, all help is appreciated :)
+83 If nn is an integer such that n≥0n≥0 then nn factorial is defined as,
n!=n(n−1)(n−2)⋯(3)(2)(1)if n≥10!=1by definitionn!=n(n−1)(n−2)⋯(3)(2)(1)if n≥10!=1by definition
Let’s compute a couple real quick.1!=12!=2(1)=23!=3(2)(1)=64!=4(3)(2)(1)=245!=5(4)(3)(2)(1)=1201!=12!=2(1)=23!=3(2)(1)=64!=4(3)(2)(1)=245!=5(4)(3)(2)(1)=120
In the last computation above, notice that we could rewrite the factorial in a couple of different ways. For instance,
5!=5(4)(3)(2)(1)4!=5⋅4!5!=5(4)(3)(2)(1)3!=5(4)⋅3!5!=5(4)(3)(2)(1)⏟4!=5⋅4!5!=5(4)(3)(2)(1)⏟3!=5(4)⋅3!
In general, we can always “strip out” terms from a factorial as follows.
n!=n(n−1)(n−2)⋯(n−k)(n−(k+1))⋯(3)(2)(1)=n(n−1)(n−2)⋯(n−k)⋅(n−(k+1))!=n(n−1)(n−2)⋯(n−k)⋅(n−k−1)!
n general, we can always “strip out” terms from a factorial as follows.
n!=n(n−1)(n−2)⋯(n−k)(n−(k+1))⋯(3)(2)(1)=n(n−1)(n−2)⋯(n−k)⋅(n−(k+1))!=n(n−1)(n−2)⋯(n−k)⋅(n−k−1)!n!=n(n−1)(n−2)⋯(n−k)(n−(k+1))⋯(3)(2)(1)=n(n−1)(n−2)⋯(n−k)⋅(n−(k+1))!=n(n−1)(n−2)⋯(n−k)⋅(n−k−1)!
We will need to do this on occasion so don’t forget about it.
Have you tried to expand the series? if you do, this is what you will get:
(0.5, 0.25, 0.1875, 0.1875, 0.234375, 0.3515625, 0.615234375, 1.23046875, 2.7685546875, 6.92138671875....etc.
So, the ratio starts with: 1/2, 3/4, 1, 1 1/4, 1 1/2.......and continues to increase by 1/4. This tells you that the series will diverge to infinity as r gets larger and larger.
