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Algebra behind the ratio test?

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]\sum\limits_{i=0}^\infty \frac{i!}{5^i}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    (1) [tex]\sum\limits_{i=0}^\infty \frac{i!}{5^i}[/tex]
    (2) [tex] = \lim_{i\rightarrow \infty}|{\frac{(i+1)!}{(5+1)^i} \cdot \frac{5^i}{i!}|[/tex]
    (3) [tex] = \lim_{i\rightarrow \infty}{\frac{(i+1)!}{5i!}[/tex]

    Could someone explain to me the jump from (2) to (3)? I can't make sense of it and my lecture notes don't mention anything in particular. Do the properties of factorials come into play here (I never learned factorials)?
     
  2. jcsd
  3. Apr 26, 2010 #2
    nothing happened, you should have

    [tex]
    = \lim_{i\rightarrow \infty}|{\frac{(i+1)!}{(5)^{i+1}} \cdot \frac{5^i}{i!}|
    [/tex]

    and the 5^i / 5^i+1 gives you 1/5
     
  4. Apr 27, 2010 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    There is no jump from (2) to (3)- Your (2) is simply wrong.

    As wizvuze said, you should have [itex]5^{i+1}[/itex], not [itex](5+ 1)^i[/itex].

    [tex]\frac{(i+ 1)!}{i!}= \frac{(i+1)i!}{i!}= i+ 1[/tex]
    and
    [tex]\frac{5^i}{5^{i+1}}= \frac{5^i}{(5^i)(5)}= \frac{1}{5}[/tex]
     
  5. Apr 27, 2010 #4
    Ok, what about:

    [tex]\frac{1}{(i+1)!} \cdot \frac{i!}{1} = \frac{i!}{(i+1)!}[/tex]

    My notes says it simplifies to this:

    [tex]\frac{1}{i+1}[/tex]

    How?
     
  6. Apr 27, 2010 #5
    [tex]\frac{i!}{(i + 1)!} = \frac{i!}{(i + 1)i!} = \frac{1}{i + 1}[/tex]
     
  7. Apr 27, 2010 #6
    Because you said that you never learned factorials let me further explain Bohrok's reply.
    5! = 5*4*3*2*1, 6! = 6*4*3*2*1 and so on.
    Think of n! as multiplying all of integers from 1 up to n together.
    i! = i*(i-1)*(i-2)...*3*2*1. The product of all the integers from 1 up to i.
    (i+1)! = (i+1)*(i)*(i-1)*(i-2)*...*3*2*1. The product of all integers from 1 up to (i+1).
    so then we have that (i+1)! = (i+1)*(i)*(i-1)*(i-2)...*3*2*1 = (i+1)*(i)!. Make this substitution on the bottom and you see that the i! on top and bottom cancel out.
     
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