Hi A.Vignesh
the nth term of the series will be (1/2n-1)*(1/2n+1)
= 1/2[(1/2n-1) - (1/2n+1)]
now we write the first three terms and check if they cancel or not
taking 1/2 common we can write the terms as 1 - 1/3 , 1/3 - 1/5 , 1/5 - 1/7
therefore on adding we see that (-1/3 1/3),(-1/5 1/5).. will get cancelled leaving behind only 1 & -1/2n+1 (as they do not get cancelled)
therefore sum is 1/2[1 - 1/2n+1)
=1/2(2n/2n+1) = n/(2n+1)
therefore the sum of the series is n/2n+1