1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another Series Question

  1. Aug 27, 2010 #1
    I know that:

    [tex]
    \sum_{n=-N}^N{1} = 2N+1
    [/tex]

    But I don't understand why.


    It seems to me that since the constant inside the summation is not dependent on n it can be moved outside of the summation, leaving nothing to sum.

    So I think the summation should actually equal one.

    Could someone help me figure out where my logic isn't right?
     
  2. jcsd
  3. Aug 27, 2010 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    If you factor out a number from a sum, like 2x+2y+2z=2(x+y+z), you divide all terms with the common factor. It has no sense to factor out "1" from the summation, as dividing by one leaves everything unchanged.The sum means that you have to add 2N+1 "1" together. It is like counting from -N to N .


    ehild
     
  4. Aug 27, 2010 #3

    berkeman

    User Avatar

    Staff: Mentor

    Even if you "move" the 1 out in front of the summation, you still leave a 1 inside. You don't leave a 0 inside. Think of the 1 inside as 1 * 1. You bring a constant 1 out in front of the summation, but that still leaves a 1 inside. Make sense?


    EDIT -- beaten out by ehild...
     
  5. Aug 27, 2010 #4

    Mark44

    Staff: Mentor

    If the summation were
    [tex]\sum_{n=1}^5{1} [/tex]
    can you see that it would be 1 + 1 + 1 + 1 + 1 = 5(1) = 5?
    What if it were [tex]\sum_{n=0}^5{1} [/tex]?
    How about this one? [tex]\sum_{n=-5}^5{1} [/tex]?
     
  6. Aug 27, 2010 #5
    [tex]\sum_{n=1}^5{1} = 5[/tex]

    [tex]\sum_{n=0}^5{1} = 6[/tex]




    [tex]\sum_{n=-5}^{5}{1}[/tex]

    [tex]=\sum_{n=-5}^{-1}{1} [/tex]

    [tex]+ \sum_{n=1}^5{1} [/tex]

    [tex]+ 1 = 11 [/tex]

    Okay. That helps a lot. Thank you all very much.
     
    Last edited: Aug 27, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook