Calculus/factorials and series
Expert: Paul Klarreich - 11/22/2005
QuestionHi, my name is leslie and I want to know how i can seperate (2x+2)! I know that if i have (x+1)! i can seperate it to be (x+1)(x)! so how do i seperate (2x+2)!
Also, i want to know how to find the sum of these series: the summation from n=0 to infinity of [ (9/8)^n +(3/4)^n ] & the summation from n=1 to infinity of [ (-1)^(n-1) (7/5)^n ]
I have tried using the geometric series formula a/(1-r) to get the sum of both of these series but for some reason im not getting the right answer. Please help me if you can. You don't know how much I will appreciate it! Thank you for your time.
Sincerely,
Leslie
AnswerOf course, the word is SEPARATE. But...
You can rewrite (2x+2)! in much the same way; whether it will do you a lot of good, I don't know.
(anything)! means to write as FACTORS all the integers from (anything) down to 1. The next integer down from (2x+2) is (2x+1), then comes (2x), (2x-1), etc. So any of these might be possible:
(2x+2)! = (2x+2)(2x+1)!
(2x+2)! = (2x+2)(2x+1)(2x)!
(2x+2)! = (2x+2)(2x+1)(2x)(2x-1)!
etc.
Also, i want to know how to find the sum of these series: the summation from n=0 to infinity of [ (9/8)^n +(3/4)^n ]
This first one is the sum of two series:
Sum (9/8)^n
n=0..inf
Sum (3/4)^n
n=0..inf
Bad news, kemo sabe. The first one diverges, so your sum does not exist.
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& the summation from n=1 to infinity of [ (-1)^(n-1) (7/5)^n ]
I have tried using the geometric series formula a/(1-r) to get the sum of both of these series but for some reason I'm not getting the right answer.
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The reason is clear: You forgot something. You think the formula for the sum of ar^n is a/(1-r). Wrong! The formula is a/(1-r) provided |r| < 1.
That little bit makes all the difference. Whenever |r| > 1, as in the first example, where r = 9/8 > 1, or in the second, where r = 7/5 > 1, the series will diverge, and the sum does not exist.
Did you miscopy the example? Were these supposed to be 8/9 and 5/7?