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Simple p-series question

  1. Jul 31, 2012 #1
    I just wanted to check that this was legal.

    [itex]\sum_5^\infty \frac{1}{(n-4)^2} = \sum_1^\infty \frac{1}{n^2}[/itex] ?
     
  2. jcsd
  3. Jul 31, 2012 #2

    Dick

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    Sure it is. They are the same series, aren't they?
     
  4. Jul 31, 2012 #3
    Thanks
     
  5. Aug 1, 2012 #4

    HallsofIvy

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    A slightly more formal derivation would be:
    [tex]\sum_{n=5}^\infty \frac{1}{(n-4)^2}[/tex]
    Let i= n- 4. Then [itex](n-4)^2= i^3[/itex] and when n= 5, i= 5- 4= 1. Of course, i= n-4 goes to infinity as n goes to infinity so
    [tex]\sum_{n=5}^\infty \frac{1}{(n-4)^2}= \sum_{i=1}^\infty \frac{1}{i^2}[/tex]

    But both "n" and "i" are "dummy" variables- the final sum does not involve either- so we can change them at will. Changing "i" to "n" in the last sum,
    [tex]\sum_{n=5}^\infty \frac{1}{(n-4)^2}= \sum_{i=n}^\infty \frac{1}{n^2}[/tex]

    The crucial point is that, as Dick said, "they are the same sequence":
    [tex]\sum_{n=5}^\infty \frac{1}{(n-4)^2}+ \frac{1}{(5-4)^2}+ \frac{1}{(6-4)^2}+ \frac{1}{(7- 4)^2}+ \cdot\cdot\cdot= 1+ \frac{1}{4}+ \frac{1}{9}+ \cdot\cdot\cdot[/tex]
    [tex]\sum_{n= 1}^\infty \frac{1}{n^2}= \frac{1}{1^2}+ \frac{1}{2^2}+ \frac{1}{3^2}+ \cdot\cdot\cdot= 1+ \frac{1}{4}+ \frac{1}{9}+ \cdot\cdot\cdot[/tex]
     
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