prove that:2^1/4 4^1/8 8^1/16 upto - Maths - Sequences and Series

Question

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

Kapil · Accepted Answer

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 ax
* ay = ax+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

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