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Math Help - Sum question

  1. #1
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    Sum, not solved yet...

    Im trying to learn something about matematical induction. This chapter starts with sums (Sigma)

    There is this easy problem I do not understand.
    It says calculate the Sigma: from k=1;to n. where (1/k-1/(k+1))

    The textbook answer says: Answer: 1-1/(n+1)

    But when I do it I get:
    starts at k=1 so I get: 1/1-1/(1+1) = 1/2

    Now I try to evaluate the last term:

    1/n-1/(n+1) /multiply by n(n+1)
    = (n+1)/n(n+1) -n/(n(n+1))
    that gives me 1/(n^2+n)

    so I get 1/2 +1/(n^2+n)
    Now to evaluate the sum I take the (first term +last term)multiply by n and devide by 2.

    That gives me (n(1/2+1(n^2+n))/2  = n/4+1/(2(n+1))

    clearly Im doing someting very very stupid...anyone got an ida whats wrong here?
    Last edited by Henryt999; February 1st 2010 at 09:09 AM.
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  2. #2
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    Quote Originally Posted by Henryt999 View Post
    Im trying to learn something about matematical induction. This chapter starts with sums (Sigma)

    There is this easy problem I do not understand.
    It says calculate the Sigma: from k=1;to n. where (1/k-1/(k+1))

    The textbook answer says: Answer: 1-1/(n+1)

    But when I do it I get:
    starts at k=1 so I get: 1/1-1/(1+1) = 1/2

    Now I try to evaluate the last term:

    1/n-1/(n+1) /multiply by n(n+1)
    = (n+1)/n(n+1) -n/(n(n+1))
    that gives me 1/(n^2+n)
    Okay, those are correct.

    so I get 1/2 +1/(n^2+n)
    Now to evaluate the sum I take the (first term +last term)multiply by n and devide by 2.
    Why? What reason do you have to think that would work? For example, if I want to sum 2+ 8+ 9+ 11, the first number is 2, the last is 11, and there are 4 numbers. Their average, (2+11)/2, multiplied by 4 is 25. but the sum is actually 30.

    Actually, I do know why you did that- for an arithmetic series that is true, and a very useful fact. But it is true only for arithmetic series, where each term is the previous term plus a fixed number. For example, 2, 8, 14, 20, ... has each number the previous number plus 6. In the example I gave, 2+ 8+ 9+ 11, the second number is 2+ 6, but the third number is 8+ 1 and the fourth number is 9+ 2. Not a constant difference.

    For your series, the first number is, as you say 1/2. The second number is 1/2- 1/3= 3/6- 2/6= 1/6 and the third number is 1/3- 1/4= 4/12- 3/12= 1/12.
    1/3= 1/2+ (-1/6) but 1/12= 1/3+ (-3/12). NOT an arithmetic series.

    That gives me (n(1/2+1(n^2+n))/2  = n/4+1/(2(n+1))

    clearly Im doing someting very very stupid...anyone got an ida whats wrong here?
    Well, not very, very; perhaps too many "very"s!
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  3. #3
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    Ahaa

    Thanks!! With your help now have one less "very". Ill try to make those verys => lim: "very stup.." =>0
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  4. #4
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    Im sorry I still dont get how to get the answer

    So I have the same issue. I posted 1.5hours ago.
    Thank you hallsofIvy but I dont grasp how to continue.

    Sigma from k=1 to n (1/k-1/(k+1))

    How on earth do I get that to be 1-1/(1+n)?

    I tried this 1/2+1/6+1/12 +1/(n-1) -1/n. Then I try to get every term that includes an n to the same denominator. so I get n/n(n-1) -(n-1)/n(n-1)

    so I get (n-n+1)/n(n-1) = 1/(n^2-n) and ofcourse that is almost very very not right.

    So well that didnt work.

    Well I tried the other way 1/2+1/6....+1/n-1/(n+1) that gives me 1/(n^2+n)

    And that doesnt help....I really dont need a complete solution maybe a hint would work wonders....
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  5. #5
    Super Member girdav's Avatar
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    You can try to write the sum for some values of n, for example until n=5.
    You will understand how to calculate that for any n.
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  6. #6
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    Induction

    Did you let k = k + 1?
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