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)