Estimate a covergent (?) infinitely nested radical

Calculus Level 5

1 5 32 288 . . . \sqrt{1\sqrt{5\sqrt{32\sqrt{288\sqrt{...}}}}}

The n n th term of the above nested radical is given by 1 n x x \displaystyle \sum_1^nx^x . Determine if the nested radical converges or diverges and than, if it converges, estimate its value k k .

8 < k < 20 8 < k < 20 Diverges 0 < k < 4 0 < k < 4 4 < k < 8 4 < k < 8

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1 solution

Cristiano Sansó
Apr 6, 2020

first of all, let's determine if it converges. We can see, the nested radical can be written as an infinite product.

n = 1 x = 1 n x x 2 n \prod_{n=1}^\infty\sqrt[2^{n}]{\sum_{x=1}^nx^{x}}

An infinite product

n = 1 a n \prod_{n=1}^\infty a_{n}

converges if

lim n a n = 1 \lim_{n \rightarrow \infty}a_{n}=1 (at least for nested radicals).

In this case, we need

lim n x = 1 n x x 2 n = 1 \lim_{n \rightarrow \infty}\sqrt[2^{n}]{\sum_{x=1}^nx^{x}}=1

in that sum, the highest value is for x = n x=n and all the other terms are negligible, so,

x = 1 n x x 2 n n n 2 n \sqrt[2^{n}]{\sum_{x=1}^nx^{x}}\sim\sqrt[2^{n}]{n^{n}}

for n n\rightarrow\infty

This is the way I solved this limit

lim n ( n n ) 1 2 n = lim n n n 2 n = lim n e n ln ( n ) 2 n = e 0 = 1 \lim_{n \rightarrow \infty}(n^{n})^{\frac{1}{2^{n}}}=\lim_{n \rightarrow \infty}n^{\frac{n}{2^{n}}}=\lim_{n \rightarrow \infty}e^{\frac{n\ln (n)}{2^{n}}}=e^{0}=1

So, now we know the product converges, let's see how to estimate the value.

let's consider the first few terms of the nested radical

1 ; 5 ; 32 ; 288 ; 3413 ; 50069 1; 5; 32; 288; 3413; 50069

so, the value of the radical (k)

k > 5 4 32 8 288 16 3413 32 50069 64 k>\sqrt[4]{5}*\sqrt[8]{32}*\sqrt[16]{288}*\sqrt[32]{3413}*\sqrt[64]{50069}

we can compare each of these to, for example, 2 n \sqrt[n]{2} so,

5 4 > 2 \sqrt[4]{5}>\sqrt{2}

32 8 > 2 \sqrt[8]{32}>\sqrt{2}

288 16 > 2 \sqrt[16]{288}>\sqrt{2}

3413 32 > 2 3 \sqrt[32]{3413}>\sqrt[3]{2}

50069 64 > 2 5 \sqrt[64]{50069}>\sqrt[5]{2}

and than

4 < 2 2 2 2 3 2 5 < k 4<\sqrt{2}*\sqrt{2}*\sqrt{2}*\sqrt[3]{2}*\sqrt[5]{2}<k

from the inequalities above, we can expect an exponential growth of n n in 2 n \sqrt[n]{2} , like

2 ; 2 ; 2 ; 2 3 ; 2 5 ; 2 7 ; 2 11 ; 2 18 ; 2 31 ; 2 54 ; 2 96 ; . . . \sqrt{2};\sqrt{2};\sqrt{2};\sqrt[3]{2};\sqrt[5]{2};\sqrt[7]{2};\sqrt[11]{2};\sqrt[18]{2};\sqrt[31]{2};\sqrt[54]{2};\sqrt[96]{2};...

To estimate a maximum value we can try to slow down the exponential growth, like

2 ; 2 ; 2 ; 2 ; 2 3 ; 2 5 ; 2 7 ; 2 10 ; 2 14 ; 2 20 ; 2 27 ; 2 38 ; 2 56 ; . . . \sqrt{2};\sqrt{2};\sqrt{2};\sqrt{2};\sqrt[3]{2};\sqrt[5]{2};\sqrt[7]{2};\sqrt[10]{2};\sqrt[14]{2};\sqrt[20]{2};\sqrt[27]{2};\sqrt[38]{2};\sqrt[56]{2};...

and now we can say

4 < k < 2 2 2 2 2 3 2 5 2 7 2 10 2 14 2 20 2 27 2 38 2 56 < 8 4<k<\sqrt{2}*\sqrt{2}*\sqrt{2}*\sqrt{2}*\sqrt[3]{2}*\sqrt[5]{2}*\sqrt[7]{2}*\sqrt[10]{2}*\sqrt[14]{2}*\sqrt[20]{2}*\sqrt[27]{2}*\sqrt[38]{2}*\sqrt[56]{2}<8

and finally

4 < k < 8 \boxed{4<k<8}

We can check with a calculator, to find out that the value of the nested radical is 6.52306176... 6.52306176...

@Cristiano Sansó , i dont understand the very first line. Can u elaborate how the nested radical can be written like that?

Priyanshu Mishra - 1 year, 2 months ago

each term of the radical multiplies all the other terms

imagine seeing it like

1 5 32 . . . = \sqrt{1} \sqrt{\sqrt{5}} \sqrt{\sqrt{\sqrt{32}}} ...=

= 1 2 1 5 2 2 32 2 3 . . . =\sqrt[2^1]{1} \sqrt[2^{2}]{5} \sqrt[2^{3}]{32} ...

Cristiano Sansó - 1 year, 2 months ago

After slowing down the exponential growth, you assumed it will be less than 8. Where is the proof for that?

Atomsky Jahid - 1 year, 2 months ago

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this is the reason I wrote estimate and not approximate... anyway when you see a growth for n

2 ; 2 ; 2 ; 3 ; 5 ; . . . 2; 2; 2; 3; 5; ...

since an exponential growth can't decrease, the next number must be at least 7, the next one at least 10, the next one at least 14, 20, 28 and so on

so we expect something much bigger than

2 ; 2 ; 2 ; 3 ; 5 ; 7 ; 10 ; 14 ; 20 ; 28 ; . . . 2; 2; 2; 3; 5; 7; 10; 14; 20; 28; ...

so considering that I slowed the exponential growth down to

2 ; 2 ; 2 ; 2 ; 3 ; 5 ; 7 ; 10 ; 14 ; 20 ; 27 ; 38 ; 56 ; . . . 2; 2; 2; 2; 3; 5; 7; 10; 14; 20; 27; 38; 56; ...

it's so unlikely that k will exceed that value.

If you compare

2 ; 2 ; 2 ; 3 ; 5 ; 7 ; 10 ; 14 ; 20 ; 28 ; . . . 2; 2; 2; 3; 5; 7; 10; 14; 20; 28; ...

2 ; 2 ; 2 ; 2 ; 3 ; 5 ; 7 ; 10 ; 14 ; 20 ; . . . 2; 2; 2; \color{#D61F06}{2}; 3; 5; 7; 10; 14; 20; ...

Cristiano Sansó - 1 year, 2 months ago

1 pending report

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