Can I use telescoping?

Algebra Level 2

$\Large \sqrt [4]{2}\times \sqrt[8]{4}\times \sqrt[16]{8}\times \sqrt[32]{16}\times \cdots = \ ?$

Give your answer to 3 decimal places.

The answer is 2.0000.

2 solutions

Vighnesh Raut
May 12, 2015

$S=\sqrt [ 4 ]{ 2 } \sqrt [ 8 ]{ 4 } \sqrt [ 16 ]{ 8 } \sqrt [ 32 ]{ 16 } .....\infty$ $S={ 2 }^{ \frac { 1 }{ 4 } }{ 2 }^{ \frac { 2 }{ 8 } }{ 2 }^{ \frac { 3 }{ 16 } }{ 2 }^{ \frac { 4 }{ 32 } }.....={ 2 }^{ x }$ $x=\frac { 1 }{ 4 } +\frac { 2 }{ 8 } +\frac { 3 }{ 16 } +\frac { 4 }{ 32 } +...\\ \frac { x }{ 2 } =\frac { 1 }{ 8 } +\frac { 2 }{ 16 } +\frac { 3 }{ 32 } +....\\ x-\frac { x }{ 2 } =\frac { 1 }{ 4 } +\frac { 2-1 }{ 8 } +\frac { 3-2 }{ 16 } +....\\ \frac { x }{ 2 } =\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\frac { 1 }{ 16 } +\frac { 1 }{ 32 } +.....$ $\frac { x }{ 2 } =\frac { \frac { 1 }{ 4 } }{ 1-\frac { 1 }{ 2 } } \\ \frac { x }{ 2 } =\frac { 1 }{ 2 } \\ x=1\\$ Hence, $S=2$

Moderator note:

Right. Did you notice that this is an Arithmetic-Geometric problem?

Nice application of AGP series.

Abhishek Sharma - 6 years, 1 month ago

Thank you....

Vighnesh Raut - 6 years, 1 month ago

Nice approach, Vighnesh. I used logarithms and a bit of differentiation to solve this, which seems silly now after looking at your solution. :)

Brian Charlesworth - 6 years, 1 month ago

Please don't use the notation $a_{1}+a_{2}+.........................................\infty$ . It implies multiplication by infinity, or something else. The infinite summation is implied.

Also, pro-tip: you don't need to put spaces put what's in curly braces - { } - and the braces. So \sqrt{x} is the exact same as \sqrt { x } :)

Jake Lai - 6 years, 1 month ago

Also, if it's just to print $\sqrt x$ , you can even ignore the braces as \sqrt x . Of course, this doesn't work if the stuff under the root is more than one symbol.

Prasun Biswas - 6 years ago

You can actually write it in a nice form as follows:

$S=2^{1/4}\cdot 4^{1/8}\cdot 8^{1/16}\cdots = \large\prod_{k=1}^\infty\left(2^k\right)^{2^{-(k+1)}}=\exp_2\left(\frac{1}{2}\cdot\sum_{k=1}^\infty\frac{k}{2^k}\right)$

Notation used : $~~~\exp_y(x)=y^{x}$

The value of the summation in the last step is easily obtained by differentiating the infinite GP sum formula w.r.t $x$ which I think is what Brian did when he said about using derivatives.

Prasun Biswas - 6 years ago

when u got x/2 = 1/4+1/8+1/16+1/32....... Then x/2=1/2(1/2+x/2) Further x= 1/2+x/2 => x=1

Raj kishore Agrawal - 5 years, 1 month ago

product 2^(n/(2^(n+1))), n=1 to infinity

WolframAlpha

Harout G. Vartanian - 4 years, 4 months ago

Adam Khan
Aug 7, 2016

Let S represent the product in question

S = 2^(1/4) * 4^(1/8) * 8^(1/16) * 16^(1/32) * ...

This product can be expressed in terms of base two:

S = 2^(1/4) * 2^2^(1/8) * 2^3^(1/16) * 2^4^(1/32) * ... = 2^ (1/4) * 2^(2/8) * 2^(3/16) * 2^(4/32) * ...

which can be simplified due to exponent laws as:

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

Lets focus for a moment on the exponent and see if we can express it as a geometric series. Let x represent this exponent.

x = 1/4 + 2/8 + 3/16 + 4/32 + ... x = 1/4 + (1/8 + 1/8) + (1/16 +1/16 +1/16) + (1/32 + 1/32 + 1/32 + 1/32) + ...

By expanding terms of x in this fashion, it becomes apparent that x can be expressed as a series of geometric series with common ratios of 1/2.

          x =  (1/4)/(1-1/2) + (1/8)/(1-1/2) + (1/16)/(1-1/2) +(1/32)(1-1/2) + ...

(1-1/2) * x = 1/4 + 1/8 + 1/16 + 1/32 + ... x/2 = (1/4)/(1-1/2) x/2 = (1/4)/(1/2) x/2 = 1/2

Therefore x = 1 and the original product can be evaluated as 2^1 or 2.

Can I use telescoping?

The answer is 2.0000.

2 solutions

Moderator note:

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