Radicals Overload

Calculus Level 3

$\large 2 \displaystyle\sqrt{2\sqrt[3]{2 \sqrt[4]{2 \sqrt[5]{2 \cdots}}}} = \,?$

Clarification : $\displaystyle e = \lim_{n\to\infty} \left(1 + \dfrac1n\right)^n \approx 2.718$ .

2

e

2^{e - 1}

2^e

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Martin Soliman
Dec 21, 2014

We solve this problem by taking the base- $2$ logarithm of the given expression. Doing so gives us $1 + \frac{1}{2} (1 + \frac{1}{3} (1 + \frac{1}{4} (1 + \frac{1}{5} (1 + \dots))))$

The above expression is equivalent to $1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \dots = e - 1$

Therefore, the expression is just $2^{e - 1}$ .

A much simpler way without using logarithms is to initially express the given expression (say $S$ ) as,

$\large{S=\displaystyle \prod_{i=1}^\infty \bigg(2^{\dfrac{1}{i!}} \bigg) \implies S=2^{\displaystyle \sum_{i=1}^\infty \bigg(\dfrac{1}{i!}\bigg)}}$

Now, using the expansion, $e=\displaystyle 1+ \sum_{i=1}^\infty \bigg( \dfrac{1}{i!}\bigg)$ we can easily rewrite $S$ as,

$\displaystyle \boxed{\Large{S=2^{e-1}}}$

P.S - I got the idea to express $S$ in that product form from Brian Charlesworth's solution in this problem. And I don't know why, but I am unable to mention him here!

Prasun Biswas - 6 years, 5 months ago

I don't think it is immediately clear why the expression S equals that infinite product. I think this must first be proved by induction or otherwise.

Toño Perez - 6 years, 5 months ago

Hi I reached your last step but didn't know what that series converged to so i used the calculator and got to 1/9! (and deduced that any further denominator bigger than 9! wont really affect the sum) which gave me 1.718281526 so I guessed its 1 below e. But i was wandering is there any way to get this without the calculator? Does it have to do with anything related to Macclaurins series or Taylor series?? Thanks :)

Farouk Yasser - 6 years, 5 months ago

Yup; the Maclaurin series for the exponential function is $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}.$ Plugging in $x = 1$ gives the desired series.

Tony Zhang - 6 years, 5 months ago

Ahhhh I see. Thanks for letting me know :)

Farouk Yasser - 6 years, 5 months ago

Radicals Overload

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