A sequence of numbers `A_nn=1,2,3` is defined as follows : `A_1=1/2` and for each `ngeq2,` `A_n=((2n-3)/(2n))A_(n-1)` , then prove that `sum_(k=1)^n A_k<1,ngeq1`

Explanation

We have,
`A_(r)=((2r-3)/(2r))A_(r-1),rge2`
`therefore2rA_(r)=(2r-2)A_(r-1)-A_(r-1)`
`rArr2rA_(r)-2(r-1)A_(r-1)=-A_(r-1)`
`thereforesum_(r=2)^(n+1)[2rA_(r)-2(r-1)A_(r-1)]=-sum_(r=2)^(n+1)A_(r-1)`
`rArr2(n+1)A_(n+1)-2A_(1)=-sum_(r=1)^(n)A_(r)`
`rArrsum_(r=1)^(n)A_(r)=2A_(1)-2(n+1)A_(n+1)`
`=1-2(n+1)A_(n+1)lt1` (as(n+1)`A_(n+1)gt0`)