Value of $y=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}}$
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)
Solution 1:
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.
Solution 2:
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 $\sqrt{4+(1+\frac{\ln 5}{3})}\approx 2,35297...$ .
Solution 3:
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
Solution 4:
$$\lim_{n\to\infty}\sqrt[n+1]{const}=1,$$ 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 $\{y_n\}$.