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I was given this problem on series by a friend.

If

$$y=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}}$$

then how to solve such problem.

I don’t want the full answer, rather, insights, mathematical facts, theorems, and relationships that would help me solve it on my own.

My efforts: I thought that the whole thing inside the square bracket must be a perfect square so we have [$4~+$ something] should be a positive perfect square but that would be like finding a trivial solution by trial and error method so I don't know how to solve it.

I also tried by squaring and checking like this $$y^2 - 4=\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}$$

so we get two factors $y-2$ and $y+2$, but still it was like same trial and error method of finding factors . So can any one help. Thanks in advance.

Edit: The only reasonable interpretation is the recurrence $y_n=\sqrt[n]{4+y_{n+1}}$ (Thanks @par)

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    $\begingroup$ Can you write a recurrence relation for $y$? $\endgroup$
    – dbanet
    Jun 29, 2016 at 17:49
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    $\begingroup$ The only reasonable interpretation is the recurrence $y_{n}=\sqrt[{n}]{4+y_{n+1}}$ so that $y_{n+1}=y_{n}^{n}-4$. The problem now is that we have no initial condition. $\endgroup$
    – parsiad
    Jun 29, 2016 at 17:50
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    $\begingroup$ Finding a closed form for expressions like this is notoriously difficult if possible at all. Take a look at some of the other similar problems (and links within) as you will see that not many of them have a known closed form solution: Q1, Q2, Q3 $\endgroup$
    – Winther
    Jun 29, 2016 at 18:04
  • $\begingroup$ Some more related question: Q4, Q5, Q6. Q6 is basically your question but with the number $2$ instead of $4$. $\endgroup$
    – Winther
    Jun 29, 2016 at 18:10
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    $\begingroup$ Searching in Approach0 returns a few posts about the same nested radical: math.stackexchange.com/questions/1782032/… and math.stackexchange.com/questions/875203/… $\endgroup$
    – Martin Sleziak
    Feb 3, 2017 at 13:38

5 Answers 5

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Too long for a comment, so

$\begin{array}\\ y &=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\ &=2\sqrt{1 + \frac14\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\ &=2\sqrt{1 + \frac14\sqrt[3]{4}\sqrt[3]{1+\frac1{\sqrt[3]{4}}\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\ &=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{\sqrt[3]{4}}\sqrt[4]{4}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4+...............}}}}\\ &=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4+...............}}}}\\ &=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4}\sqrt[5]{1+...............}}}}\\ &=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{4^{1/20}}\sqrt[5]{1+...............}}}}\\ \end{array} $

It looks like there is a pattern of $\dfrac1{4^{1/n-1/(n+1)}} =\dfrac1{4^{1/(n(n+1))}} $ which might make it easier to get a more solvable recurrence.

And, of course, $4$ can be replaced by any value, probably preferably a square.

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    $\begingroup$ I don't understand what to get out of your last line with the recurrence, as it is merely a true statement. $\endgroup$
    – Simply Beautiful Art
    Jul 1, 2016 at 13:32
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This is only a hint, because the following argumentation is a rough calculation.

$y^n_n=4+y_{n+1}$

If we assume a value $>1$ for $y^n_n$ we can set $y_n\approx 1+\frac{a}{n}$ for large $n$ so that we get

$(1+\frac{a}{n})^n\approx 4+(1+\frac{a}{n+1})$ which means for $n\to\infty $ the equation $e^a=5$ or simply

$a=\ln 5\approx 1.6094379...$ .

It follows (for small $n$, here $n:=2$) $\enspace y:=y_2\approx \sqrt{4+(1+\frac{\ln 5}{2})}\approx 2,409298...$ .

I think this value is a good upper bound for $y$ .

A lower bound should be $\sqrt{4+(1+\frac{\ln 5}{3})}\approx 2,35297...$ .

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I got with high precision $$y \approx 2.401615526026297355671587971656109034017...$$ when started with any value $x_0$ between $10$ and $1e20$ and the root-index of 1000.
The simple continued fraction of the approximation having 793 terms was (reading from left to right and then top-down):

    2   2    2   24   2     1    3    1   1     5
    1   1    7    1   2     1    1    4   3     1
    1   1   15  274   2    49    1    1   1    16
    2   3    1    2   1    48   10    4   1    14
    7   1    2   19  27     3    4    1   8    11
   20   1    5    3   4     1   52    1   2     1
    2   8    3    2   2     1   60    1   1     4
    3   8    2    1  69     1    5    1   3    10
   10   2    1   21  12     1    4    1  15     2
    3   1    1    9   3     8    1    1   1     2
    6   1  100    4  11     3    1    3   6     1
    1   4    3    2   2    16    2    5   2     6
    2   1    2   11   6     2   40    1   1     8
    1   5    1    1   3     2    7    1   1    15
    3   1    2    2   2     1    1    1   4     7
    1   1    6    3   2    12   13  890   1     3
    2   1    3    7   2    14  563    3  36     2
    3   9    7    1   1     1    3    1   2    13
    9   2    1    3   2     2    7   12  11     2
    1   4    1    1   1     7    5    1   4     5
    1   5    1   60   1     2    3    1  14     1
    1   6    1    1   1     4    2   59   1     2
    1  26   41    1   6     6    1    1   1     1
    2   1    2    2   1     4    1   12   1     2
    2   2   20    5   4     8    6    1   5     1
    2   1    4    2   3     2    2    2   3     3
    1   4    2    1   1    15   27    1   2     1
    1   1    3    1   4     1    4    1   1     6
   18   9    1   23   3     1    1    2  11     1
    1  24    1    5   1     1    2   13   1    22
    6   1    8    1   3     1    3    1   4     2
    1  33    4    1   2     1    2    1   2     8
  170   2    2    6   6    17    1    2   1     2
    2   6    1    1   2     1    2    1   4    13
    1   2    1    2   3     1    1    3   2     1
    2   6    5    1   2    15    2    2  13     1
   19   1    2    3   1     1    1    5   3     1
   28   1    3    1   2     1    1    2   5     1
    3   2    1    2   2     1   22    2   2     4
    1  21    2    1   9     1    7    1   6     7
    5   1   18    4   2     2    1    2   1     1
    1   1    3    1   1     1    6   27  18     1
    6   1    9    1   4     3    1   28   2     8
    6   1   35    1   1     2   56    1   1     1
    2   1    1    7   1     1    4    2   1     2
    1   1    1    3   1     1   37    1   8  3154
    1   1    3    1   3     1   11    1   9     1
    3   2    7    2   1    56    2   11   8     1
    2   1    1    2   1     4   16    1   2     3
    1   1    2    1   1    15    1    1   1     1
  254   1    1    7   1     4    1    2   4     1
    2   1    1    1   1     1    2    2   1     2
    1  11    5    2   1    89    1   10   1     1
    3   3    1   11   1     1    1    1  13     1
    5  37    1    6   1     7    1   11   1     2
    6   2    2    4   2     1    2    1   4     3
    3   4    7    1   7     1    3   14  17     2
    1   4    1    4   3     1   80    6   1     4
    1   2    1    9   5     1    1    1   7     3
    2   3    1    1  87     1    1   12   1   177
    1   2    1    1   2     6    1   47   7     7
    2   4    1    6   1     2    2    1   4     2
    1   4    1    3   4     4   69    1  10     5
    1   4    1    3   4     1   14    3   1     1
    2  11    1    1   3     1    2    1   2     1
   10   2    1    1  11    25    2    6   3    13
    1  12   12    4   1     4    4    1   3     1
    1   6    1   11   3     5    1    1   1     4
    2   3    1   11   4     2    1    1   1     1
    4   4    2    1   4     4  190    2  20    14
    1   5    1   14   1     1    1    2   2     1
    1   5    2    1   1  2289    2   14   6    37
    1   5    1   21   2    16    1    9   2     2
   11  13    6    1   2     4    3    6   3    55
    1   1   10    1   2    30    4   14   6     1
    3   2    1    1   2     1    1    2   1     1
   13   1    1    5   1     2    1    1   2    17
    3   2    2    1   1     5    2    5   2     3
    1   1    1    1   7     2    1    1   1     2
   ...

A rough procedure to test is the following in Pari/GP. Assume $\small x_1=1$ as lower bound, $\small x_2=1e200 $ as upperbound, and some initial root-index, say 30. Then compute the nested radical down to index 2:

index=30;x1=1;x2=1e200
forstep(k=index,2,-1, x1=(4+x1)^(1/k); x2=(4+x2)^(1/k); print([x1,x2,x2-x1]) );

Output: 

   x1(incr.)       x2(decr.)        difference
 ------------------------------------------------
  1.05511306354  4641588.83361           4641587.77850
  1.05746641188  1.69779429681          0.640327884932
  1.05959637667  1.06411735764        0.00452098097191
  1.06188715066  1.06192227795      0.0000351272888663
  1.06436097020  1.06436125429    0.000000284083700916
  1.06704070281  1.06704070521  0.00000000239421214644
  1.06995316160  1.06995316162       2.10650144506E-11
  1.07313002353  1.07313002353       1.93857373231E-13
  1.07660898434  1.07660898434       1.87000018397E-15
  1.08043525836  1.08043525836       1.89516657781E-17
  1.08466355021  1.08466355021       2.02307283136E-19
  1.08936068075  1.08936068075       2.28122116120E-21
  1.09460913536  1.09460913536       2.72577952262E-23
  1.10051193561  1.10051193561       3.46358372476E-25
  1.10719944779  1.10719944779       4.69913367993E-27
  1.11483908939  1.11483908939       6.83842220013E-29
  1.12364948065  1.12364948065       1.07306679076E-30
  1.13392160888  1.13392160888       1.82678325796E-32
  1.14605141776  1.14605141776       3.39829163458E-34
  1.16059171388  1.16059171388       6.96744109268E-36
  1.17833818093  1.17833818093       1.59090319833E-37
  1.20047878604  1.20047878604       4.09793745817E-39
  1.22886851186  1.22886851186       1.21042372084E-40
  1.26657139203  1.26657139203       4.18852719443E-42
  1.31902996773  1.31902996773       1.74838380889E-43
  1.39690376355  1.39690376355       9.18334334489E-45
  1.52417968370  1.52417968370       6.48384053520E-46
  1.76775713485  1.76775713485       6.91617339503E-47
  2.40161552603  2.40161552603       1.43990020886E-47
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    $\begingroup$ I tried finding a minimal polynomial for such an approximation, but the best I found is the following which should be good up to 122 digits but won't match after that: $$8258028589600 y^{8} - 10398489573954 y^{7} - 16164983819372 y^{6} - 8219788311584 y^{5} - 12767863340898 y^{4} - 11126864974160 y^{3} + 3533772609633 y^{2} - 17611448551760 y + 11820477102949$$ $\endgroup$
    – MvG
    Feb 9, 2017 at 17:27
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$$\lim_{n\to\infty}\sqrt[n+1]{const}=1,$$ so $$y=\sqrt{4+\sqrt[3]{4+\sqrt[4]{\dots+\sqrt[n]{5+o(1)}}}} = \lim_{n\to\infty} y_n,$$ where $$y_n=\sqrt{4+\sqrt[3]{4+\sqrt[4]{\dots+\sqrt[n]{5}}}}.$$ Using calculator, one can get: $$y_2 = \sqrt 5 \approx 2,2360679774997896964091736687313,$$ $$y_3=\sqrt{4+\sqrt[3]5} \approx 2,3895555960631460037703850634224,$$ $$y_4=\sqrt{4+\sqrt[3]{4+\sqrt[4]5}}\approx 2,4009740604119511706503467195743,$$ $$y_5=\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{5}}}}\approx 2,4015885636119483509721951488115,$$ $$y_6=\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\sqrt[6]{5}}}}}\approx 2,4016145888373557072804406968762\dots,$$ indicating a good practical convergence of the sequence $\{y_n\}$.

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The question is pretty close to this one, just replace $\,2\,$ by $\,4\,$ in:

There is a little (Delphi Pascal) computer program in one of the answers. At the end of that piece of code we find:

begin
  anatoly(2,13); { at double precision }
end.
Replace this by:

begin
  anatoly(4,13); { at double precision }
end.
To get:

 4.00000000000000E+0000  2.40161552602630E+0000 +/- 1.80950091432944E-0018
Which is - no big surprise - in concordance with the numerical value as given by Gottfried Helms in his answer .

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