Question Number 10493 by ABD last updated on 14/Feb/17


Answered by mrW1 last updated on 14/Feb/17

since (n/((n+1)!))=(1/(n!))−(1/((n+1)!))  (1/(2!))+(2/(3!))+(3/(4!))+(4/(5!))+...+((17)/(18!))  =((1/(1!))−(1/(2!)))+((1/(2!))−(1/(3!)))+((1/(3!))−(1/(4!)))+∙∙∙+((1/(17!))−(1/(18!)))  =(1/(1!))−(1/(18!))  =1−(1/(18!))

Answered by robocop last updated on 14/Feb/17

    a1=1/2!=1/2  an=17/18!=0  d=((2×2!−3!)/(2!3!))= −1/6  n=((an−a1)/d)+1= 4    Sn=(((a1+an))/2)×n  Sn=1

Commented byFilupS last updated on 15/Feb/17

this method wont work because a sequence  of factorials is product based and they  grow exponentially larger. i.e.    let S=n!+(n+1)!+(n+2)!+...  if d=t_2 −t_1 =t_3 −t_2   ∴ d=(n+1)!−n!=(n+2)!−(n+1)!         =n!∙((n+1)−1)=(n+1)!∙((n+2)−1)         =n!∙(n)=(n+1)!∙(n+1)         =n!∙(n)=(n)!∙(n+1)^2   this is a clear contradiction, because the  RHS is larger by a factor of  (((n+1)^2 )/n),   rather than being larger than a common  difference.

Commented byrobocop last updated on 15/Feb/17


Commented byFilupS last updated on 16/Feb/17

incorrect     ((18!−1)/(18!))≠1  (a/b)=1⇔a=b  18!−1≠18!  ∴((18!−1)/(18!))≠1