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.
( ∑ [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.