07 Oct '15 08:244 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.