- 07 Oct '15 08:24 / 4 editsI 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. - 08 Oct '15 03:00You 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) - 08 Oct '15 07:11 / 2 edits

Thanks for that! That's brilliant! I understood that the moment I read it.*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)

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. - 08 Oct '15 17:22

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:*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.

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.