prove that:2^1/4 . 4^1/8 . 8^1/16............upto infinity=2

Let Z =2^1/4 * 4^1/8 * 8^1/16 ..............upto infinity

Z = 2^1/4 * (2^2)^1/8 * (2^3)^1/16 .................upto infinity

Z = 2^1/4 * 2^2/8 * 2^3/16 ...............up to infinity

Z = 2^(1/4 + 2/8 + 3/16 ..............upto infinity) ...........(1) (Using a^{x} * a^{y }= a^{x+y} )

Now let

S = 1/4 + 2/8 + 3/16 + 4/32 ........................upto infinity ...........(2)

[Thus, equation (1) becomes Z = 2^S .........(A)]

Multiplying both sides by 1/2 we get ( we multiply by 1/2 as the denominators form a Geometric Progression with common ratio 2 the numerators form an Arithmetic Progression with common difference 1)

1/2 S = 1/8 + 2/16 + 3/32 + ..............upto infinity ................ (3)

Subtracting equation (3) from equation (2) we get

S - 1/2 S = (1/4 + 2/8 + 3/16 + 4/32 .........upto infinity) - (1/8 + 2/16 + 3/32 + ..............upto infinity)

= 1/2 S = 1/4 + (2/8 - 1/8) + (3/16 - 2/16) + (4/32 -3/32) ..............upto infinity (Grouping like terms together)

= 1/2 S = 1/4 + 1/8 + 1/16 + 1/32 + ................upto infinity

Which is a Infinite Geometric Progression with

First term a = 1/4 common ratio r = (1/8)/(1/4) = (1/8) * (4/1) = 1/2

Thus , 1/2 S = (1/4)/(1 - 1/2) [As sum of infinite G.P. = a/(1 - r)]

= 1/2 S = (1/4) / (1/2)

= 1/2 S = (1/4) * (2/1) = 1/2 S = 1/2

= S = (1/2) / (1/2)

= S = 1

Putting value of S in equation (A)

Z = 2^1 = 2 .

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