Re:部分分数分解の真実
1/(k*(k+1)*(k+2))=(1/2)*(1/(k*(k+1))-1/((k+1)*(k+2)))
1/(k*(k+1)*(k+2)*(k+3))=(1/3)*(1/(k*(k+1)*(k+2))-1/((k+1)*(k+2)*(k+3)))
...
1/(k*(k+1)*(k+2)*...*(k+m))=(1/m)*(1/(k*(k+1)*...*(k+m-1))-1/((k+1)*(k+2)*...*(k+m)))
なので、
1/(1*2*3)+1/(2*3*4)+...+1/(n*(n+1)*(n+2))=(1/2)*(1/(1*2)-1/((n+1)*(n+2)))
1/(1*2*3*4)+1/(2*3*4*5)+...+1/(n*(n+1)*(n+2)*(n+3))=(1/3)*(1/(1*2*3)-1/((n+1)*(n+2)*(n+3)))
...
1/(1*2*3*...*(m+1))+...+1/(n*(n+1)*(n+2)*...*(n+m))=(1/m)*(1/m!-n!/(n+m)!)
というのもいえますね。
