What an Amusing Nested Radical!

Calculus Level 3

$\Large \sqrt{2\sqrt[3]{2\sqrt[4]{2\sqrt[5]{2\cdots}}}}$

Let the value of the above expression be denoted by $I$ . Find $\lfloor1000I\rfloor$ .

Details and Assumptions:

You may use a scientific calculator to evaluate the final result.

The answer is 1645.

1 solution

Micah Wood
Mar 17, 2015

$\large \sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\cdots}}}}$ can be rewritten as

$\Large \begin{aligned} ~x^{\frac{1}{2}(1+\frac{1}{3}(1+\frac{1}{4}(1+\frac{1}{5}(1+\cdots))))} &= x^{\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\cdots} \\~\\&=x^{e-2} \end{aligned}$

since $\displaystyle e = 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\cdots$ .

Let $x=2$ and we have $I = 2^{e-2} = 1.6452\ldots$

So we have $\lfloor1000I\rfloor=\boxed{1645}$

this question and your solution increased my knowlede, thanks for sharing knowledge.

Maroof Ali - 6 years, 2 months ago

After a lot struggling to find the answer , I am really delighted by ur solution @Micah Wood

Kunal Gupta - 6 years, 2 months ago

amazing, tks!

Triet Tran - 6 years, 2 months ago

@Micah Wood , Nice solution, but it would help if you would derive the sum to be $e-2$ for most of the users. Thanks!:)

A Former Brilliant Member - 6 years, 2 months ago

I'll give you a clue, consider the taylor series of $e^x$ and set $x=1$ .

Julian Poon - 6 years, 2 months ago

@Julian Poon – Haha...Thanks @Julian Poon , i knew the derivation! I was just saying that it would be important for someone who doesn't:D

A Former Brilliant Member - 6 years, 2 months ago

@A Former Brilliant Member – Oh....

Yeah, I agree with you. It's kinda important.

Julian Poon - 6 years, 2 months ago

@Micah Wood really amazing........you are genius "Micah Wood"

Nabil Rafiq - 6 years, 2 months ago

Beautiful... (removes tear from corner of eye)

Pieter Breughel - 4 years, 8 months ago

This is amazing. Solutions like this exemplify why I love mathematics. Thank you

David Orrell - 4 years, 7 months ago

Can you elaborate how you obtained $\large 2^{\frac{1}{2} (1+\frac{1}{3} (1+\frac{1}{4} (1+\frac{1}{5} (1+\cdots))))}$ from $\large \sqrt{2\sqrt[3] {2\sqrt[4]{2\sqrt[5] {2\cdots}}}}$ ??

Akshat Sharda - 5 years, 8 months ago

$\sqrt { 2\sqrt [ 2 ]{ 2\sqrt [ 3 ]{ 2... } } } =2^{ \frac { 1 }{ 2 } }\cdot 2^{ \frac { 1 }{ 2 } \cdot \frac { 1 }{ 3 } }\cdot 2^{ \frac { 1 }{ 2 } \cdot \frac { 1 }{ 3 } \cdot \frac { 1 }{ 4 } }...=2^{ \frac { 1 }{ 2! } }\cdot 2^{ \frac { 1 }{ 3! } }\cdot 2^{ \frac { 1 }{ 4! } }...=2^{ \frac { 1 }{ 2! } +\frac { 1 }{ 3! } +\frac { 1 }{ 4! } ... }$

Julian Poon - 5 years, 8 months ago

Didn't think of nesting in this way can be simplified. This is a very good "question and answer".

Lu Chee Ket - 5 years, 8 months ago

Same way!!

Dev Sharma - 5 years, 7 months ago

Please forgive my ignorance, but shouldn’t the solution equal 1000 because e = 2 mathematicly, therefore 2 to the e – 2 power would be the same as 2 to the 0 power. Then I would equal 1 and 1000I would equal 1000. That’s how I solved it after I got the equation with the denominator factorially increasing in value.

Fresh 750 - 3 years, 7 months ago

No, the value of e is approximately 2.71828 . My solution relies on the knowledge of Taylor series of $e^x$ at $x = 1$ .

Micah Wood - 3 years, 7 months ago

Totally over my head :-)

Conrad Winkelman - 2 years, 5 months ago

i've done the same, we are great!

Andrea Virgillito - 4 years, 9 months ago

same solution (y) :)

Moaaz Al-Qady - 5 years, 10 months ago

What an Amusing Nested Radical!

The answer is 1645.

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