Solve please.

•3^{1/2x31/4x31/8x31/16x…...........}upto infinty
•3^{1/2 +1/4+1/8+1/16 +1/32 …...........}
•Now, 1/2 +1/4+1/8+1/16 +1/32+... is an infinite series with
common ratior= (1/4)/(1/2)= (1/2)

•So, the sum of an infinite GP with r<1 is S(inf.)= a/(1-r)

•Hence, Series S:3^{1/2 +1/4+1/8+1/16 +1/32 …..........} = 3^{[(1/2)/(1-r)]}
•S= 3^{[(1/2)/(1-(1/2))]}
•S= 3^[1]
•S=3