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  1. 07 Oct '15 08:24 / 4 edits
    I accidentally, via a numerical approach in a computer program, discovered, for many combinations of h and x values, that the cumbersome expression:

    ( ∑ [X = h, x] 1/(X(X+1)) ) / ( ∑ [X = h, ∞] 1/(X(X+1)) )

    where h ∈ ℕ, x ∈ ℕ, h ≠ 0, x ≠ 0, appears to simplify to just:

    1 – ( h / (x + 1))

    But, to put in my book I am writing, I want the algebraic proof of this simplification. But the problem is I cannot see why this simplification appears to work.
    So, the mystery I want to solve is the algebraic reason why:

    ( ∑ [X = h, x] 1/(X(X+1)) ) / ( ∑ [X = h, ∞] 1/(X(X+1)) ) = 1 – ( h / (x + 1))

    where h ∈ ℕ, x ∈ ℕ, h ≠ 0, x ≠ 0,

    any ideas?

    Don't know if this helps but I have already worked out that the denominator of that fraction can be expressed as:

    ∑ [X = h, ∞] 1/(X(X+1)) = 1 – ∑ [X = 1, h – 1] 1/(X(X+1))

    which at least gets rid of that unwelcome "∞".

    Also note that;
    1 – ( h / (x + 1)) = (x + 1 – h) / (x + 1)
    if that is of any help.
  2. Standard member DeepThought
    Losing the Thread
    08 Oct '15 03:00
    You can rewrite your typical term, 1/n*(n+1) as 1/n - 1/(n + 1) using partial fractions, this gives the method to solving your sum:

    sum {h, g} 1/n*(n+1) = sum {h,g} 1/n - 1/(n+1)
    = [1/h + 1/ (h + 1) + ... + 1/(g - 1) + 1/g] - [1/(h + 1) + 1/(h + 2) + ... + 1/g + 1/(g + 1)]
    = 1/h - 1/(g + 1)

    Your quotient is: Q = S1/S2 = sum {h, x} 1/n(n+1) / sum{h,infy} 1/n(n+1). Using my formula:

    S1 = sum {h,x} 1/n*(n+1) = 1/h - 1/(x + 1)

    and

    S2 = limit [x -> infty] S1 = 1/h

    Q = S1/S2 = 1 - h / (x + 1)
  3. 08 Oct '15 07:11 / 2 edits
    Originally posted by DeepThought
    You can rewrite your typical term, 1/n*(n+1) as 1/n - 1/(n + 1) using partial fractions, this gives the method to solving your sum:

    sum {h, g} 1/n*(n+1) = sum {h,g} 1/n - 1/(n+1)
    = [1/h + 1/ (h + 1) + ... + 1/(g - 1) + 1/g] - [1/(h + 1) + 1/(h + 2) + ... + 1/g + 1/(g + 1)]
    = 1/h - 1/(g + 1)

    Your quotient is: Q = S1/S2 = sum {h, x} 1/n(n+1) / sum{ ...[text shortened]... +1) = 1/h - 1/(x + 1)

    and

    S2 = limit [x -> infty] S1 = 1/h

    Q = S1/S2 = 1 - h / (x + 1)
    Thanks for that! That's brilliant! I understood that the moment I read it.
    I just never thought of the simple idea of turning that into partial fractions and then using that to cancel down the terms and, as a result, the problem had me totally stumped for ages.

    I will definitely put that proof in my book. I will gladly cite credit to you in my book for that if you like? I am going to be doing that a lot for other people in many parts of my book anyway as my book inadvertently partly builds on top of other people's work quite a bit although it will also have a great deal that is purely my own insight.
  4. Standard member DeepThought
    Losing the Thread
    08 Oct '15 17:22
    Originally posted by humy
    Thanks for that! That's brilliant! I understood that the moment I read it.
    I just never thought of the simple idea of turning that into partial fractions and then using that to cancel down the terms and, as a result, the problem had me totally stumped for ages.

    I will definitely put that proof in my book. I will gladly cite credit to you in my book for ...[text shortened]... people's work quite a bit although it will also have a great deal that is purely my own insight.
    You can also do it by iteration. Let S(h,x) = sum{h,x} 1/n(n+1) and suppose that your formula works for all h and all x up to some value y > h. We need to show that it is also correct for S(h, y + 1). We have:

    S(h, y) = 1/h - 1/(y + 1).

    S(h, y + 1) = sum{h , y + 1} 1/n(n+1) = sum{h,y} 1/n(n+1) + 1/(y + 1)(y + 2) = S(h, y) + 1/(y + 1)(y + 2)

    Substituting the formula for S(h,y) we get:

    S(h, y + 1) = 1/h - 1/(y + 1) + 1/(y + 1)(y + 2) = 1/h - [(y + 2) - 1]/(y + 1)( y + 2) = 1/h - 1/(y + 2).

    This shows that if S(h, y) is given by 1/h - 1/(y + 1) then the formula will also give the correct result for S(h, y + 1).

    S(h, h) = 1/h(h + 1) = 1/h - 1/(h + 1) so the formula works for the case S(h, h) for any h in the set of positive definite integers. This completes the proof.