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Determine whether the series converges or diverges?
We have the series (2^k)k! / (k+2)! from k = 1 to Inf.
I realize the method is to rewrite the factorials in such a way that the expression becomes more simple and can then have a test ran on it, but I was not taught how to do this and I cannot find a simple explanation for it online.
Thanks. I don't need anyone to do any convergence tests, just showing me how to simplify the expression would be nice.
3 Answers
- 1 decade agoFavorite Answer
Allrighty! Remember k!= k* k-1 * k-2 * ... * 3*2*1
ex: 5! = 5*4*3*2*1
Now look at (k+2)! = k+2 * k+1 * k * k-1* ... * 2*1
Now consider k+2 * k+1 * [k * k-1 * k-2 * ... 2*1]
Does the quantity inside the [ ] look similar to the top line?
You can now rewrite it as k+2 * k+1 * k!
Ex: 7! = 7*6*5*4*3*2*1 and 7=5+2 so
(5+2)! = (5+2)*(5+1) * 5*4*3*2*1 and just like the second line we know 5!=5*4*3*2*1 so
7! = (5+2)! = (5+2)*(5+1)*5!
Now for your series
look at the part k! / (k+2)! we now know k! / [(k+2) * (k+1) * k!] and we can cancel the k!.
Thus your series simplifies to (2^k) / [(k+2)(k+1)] from k = 1 to inf.
- 1 decade ago
Simplification of a factorial is very simple. Consider n!:
n! = (n)(n-1)(n-2)...(1)
in the same way, (k+2)! = (k+2)(k+1)(k)(k-1)...(1).
Ignoring the first two binomials, take (k)(k-1)...(1)... what does that look like?
k(k-1)...(1) = k!
So, that bottom term becomes (k+2)(k+1)k!.
The k! cancels out on both numerator and denominator. This leaves you with a good bit simpler expression:
(2^k) / (k+2)(k+1)
This should be easier to work with.
Source(s): lots of advanced college math - 1 decade ago
Well, k!=k*(k-1)*(k-2)*......3*2*1 and (k+2)!=(k+2)*(k+1)*k*(k-1)*......*3*2*1=(k+2)*(k+1)*k!
So, we may cancel out the k! on the top, and that leaves the bottom with (k+2)*(k+1).
This makes the series (2^k)/((k+1)*(k+2)), which you can presumely test on now.
Alternatively, you could have just applied the ratio test to the original puppy that had the factorials and all.
