# What is $1+(2+(3+(4+…+(n-1+n^{1/n})^{1/(n-1)}) …)^{1/4})^{1/3})^{1/2}$ equal to? [closed]

The correct result should be somewhere close to $2.9116$, the problem is: what is the exact formula to calculate this?

$$\lim_{n \to +\infty} 1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\cdots \sqrt[n]{n}}}}$$

Sorry the correct question is: is there a closed form representation for this?

• Isn't the definition an exact formula? Or: what are the reasons for expecting a somewhat simpler expression? – Jack D'Aurizio Jan 19 '17 at 20:47
• I don't know if there is a definition for this or not , I haven't heard of it. – shadow_wxh Jan 19 '17 at 20:52
• So you have $a_n = 1 + \sqrt{2+\sqrt[3]{3+\cdots +\sqrt[n-1]{(n-1) +\sqrt[n]{n}}}}$, and want to know $\lim_{n\to\infty}a_n$? – MPW Jan 19 '17 at 21:09
• This is the exact formula. Even if there was a shorter one (there is not), this still would be an exact formula – Yuriy S Jan 19 '17 at 22:14
• I removed the "sum of" from the title as you seem not to consider a sum of the expressions given but rather an expression that happens to be a sum. – quid Jan 19 '17 at 22:24

Just to give something substantial an an answer, I will provide a Mathematica code for computing this number with high precision.

Nm=20;
A=Table[j,{j,1,Nm}];
V=Table[j,{j,1,Nm}];
Do[v0=A[[n]]^(1/n);
Do[v1=(A[[n-j]]+v0)^(1/(n-j));
v0=v1,{j,1,n-1}];
V[[n]]=v0,{n,1,Nm}];
Print[MatrixForm[N[V,30]]]
1.00000000000000000000000000000
2.41421356237309504880168872421
2.85533004349830124296470905097
2.90798456765468398163351993468
2.91148304056081180846709800043
2.91163449677407059738451903816
2.91163911038987173661011302461
2.91163921441793019894893209160
2.91163921622081710315790195458
2.91163921624554688768942416974
2.91163921624582174103946188723
2.91163921624582426431140283962
2.91163921624582428375510464360
2.91163921624582428388258648467
2.91163921624582428388330599114
2.91163921624582428388330952238
2.91163921624582428388330953758
2.91163921624582428388330953764
2.91163921624582428388330953764
2.91163921624582428388330953764


The first $100$ digits of the number are:

2.911639216245824283883309537640624664608581539406223824296679924425963256451757852869841727384105859

The continued fraction representation starts with:

[2; 1, 10, 3, 6, 1, 1, 3, 12, 1, 12, 4, 1, 1, 22, 3, 1, 1, 2, 1, 1, 35, 1, 4, 3, 14, 1, 1, 1, 2, 1, 6, 3, 1, 13, 33, 1, 1, 14, 1, 12, 81, 1, 16, 22, 1, 5, 4, 83, 1, 5, 1, ...]

$$2.9116392\ldots=2+\cfrac{1}{1+\cfrac{1}{10+\cfrac{1}{3+\cfrac{1}{6+\dots}}}}$$

The Egyptian fraction representation of the fractional part by the greedy algorithm starts with:

$$2.9116392\ldots=2+\frac{1}{2}+\frac{1}{3}+\frac{1}{13}+\frac{1}{724}+\frac{1}{628726}+\frac{1}{545647113510}+\dots$$

WolframAlpha, OEIS or Inverse Symbolic Calculator do not recognize the digits of this number.

BTW, this is what I do when I come up with some number without a known closed form. I compute it in different representations and try to see if anything interesting comes up. Nothing here so far.

Now regarding the actual limit of that particular formula, I do not know whether it corresponds to some "absolutely well defined, yet easily written down, value", like a given logarithm, root or whatever.

Srinivasa Ramanujan is also well-known notably for his remarkable work using nested radicals.

And this question and its answer has extensive information on the subject as well.