# Closed form for 1/(k x (k+1) x (k+2))?

Σ 1/(k x (k+1) x (k+2)) ; k goes from 1 to n

I need a proof for this as well..

Relevance

Decomposing 1/[k(k+1)(k+2)], you have:

A/k + B/(k+1) + C/(k+2) = 1/[k(k+1)(k+2)] for some numbers A,B,C. Therefore:

A(k+1)(k+2) + B(k)(k+2) + C(k)(k+1) = 1

which must hold true for all valid values of k. To determine A,B,C, plug in different values of k.

If k = 0, you have A=1/2.

If k = -1, B = -1

If k = -2, C = 1/2

So Σ 1/[k(k+1)(k+2)] = Σ (1/2)/k + (1/2)/(k+2) - 1/(k+1).

This is a telescoping series.

The terms for Σ (1/2)/k are (1/2)/1, (1/2)/2, (1/2)/3,... (1/2)/n.

The terms for Σ (1/2)/(k+2) are (1/2)/3, (1/2)/4,... (1/2)/(n+2).

The terms for Σ -1/(k+1) are -1/2, -1/3, -1/4... -1/(n+1).

The terms in the first 2 series can be combined to cancel out the same term in the 3rd series. The terms cancels out from 3 to n.

So what's left of the first series are (1/2)/1 and (1/2)/2.

What's left of the 2nd series are (1/2)/(n+1) and (1/2)/(n+2).

And what's left of the 3rd series are -1/2 and -1/(n+1).

(1/2)/1 + (1/2)/2 + (1/2)/(n+1) + (1/2)/(n+2) - 1/2 - 1/(n+1)

= 1/4 + (1/2) [1/(n+2) - 1/(n+1)].

Using partial fractions, this becomes:

Σ [1/2k - 1/(k+1) + 1/2(k+2)]

= ½ Σ 1/k - Σ 1/(k+1) + ½ Σ 1/(k+2)

=½ [1+1/2+1/3+...+1/n] - [1/2+1/3+...+1/(n+1)] + ½[1/3 + 1/4 +...+ 1/(n+2)]

much of the middle terms cancel... leaving...

= 1/2 - 1/4 - 1/{2(n+1)} + 1/{2(n+2)}

= [n^2+3n+1] / [4(n+1)(n+2)]

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