If `a_(1)=1,a_(n+1)=(1)/(n+1)a_(n),a ge1`, then prove by induction that `a_(n+1)=(1)/((n+1)!)n in N`.

Explanation

Let `P(n):a_(n+1)=(1)/((n+1)!),n in N`.....(i)
where `a_1=1 and a_(n+1)=(1)/((n+1))a_(n),n ge 1` ......(ii)
Step I For n=1, form Eq. (i) , we get `a_(2)=(1)/((1+1)!)=(1)/(2!)`
But from Eq. (ii) , we get a_(2)=(1)/((1+1)),a_(1)=(1)/(2)(1)=(1)/(2)`
which is true ,
Also, for n=2 from Eq. (i) we get `a_3=(1)/(3!)=(1)/(6)`
But from Eq. (ii) , we get `a_3=(1)/(3),a_2=(1)/(3).(1)/(2)=(1)/(6)`
which is also true .
Hence ,P(1) and