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Nice infinite series (Posted on 2023-04-11) Difficulty: 3 of 5
Let
S=1+sqrt(2+sqrt(3+sqrt(4+…. ...n-1 +sqrt( n))))

What is the limit of S when n approaches linfinity?

Attributed to Ramanujan

No Solution Yet Submitted by Ady TZIDON    
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Solution computer calculation and some observations | Comment 1 of 3
for upto=400:100:500;
  tot=vpa(0);
  for n=upto:-1:2
    tot=sqrt(tot+vpa(n));
  end
  S=1+tot
end

finds

>> niceInfiniteSeries
S =
3.0903275767905763591925445066881
S =
3.0903275767905763591925445066881

showing the same value whether we go out to n=400 or n=500.

Of course the value shown is accurate only to this degree of precision.

It actually reaches this value, to this degree of precision, at n=35:

for upto=2:37
  tot=vpa(0);
  for n=upto:-1:2
    tot=sqrt(tot+vpa(n));
  end
  S=1+tot;
  disp([upto S])
end

>> niceInfiniteSeries
[2, 2.4142135623730950488016887242097]
[3, 2.9318516525781365734994863994578]
[4, 3.0581710272714922503219810475805]
[5, 3.084373625265516202761085111978]
[6, 3.0893027111184985194327248493655]
[7, 3.0901619447518881524805789618568]
[8, 3.0903022659516802174745853102976]
[9, 3.0903238988326671788247508659392]
[10, 3.0903270662019340002047552811406]
[11, 3.0903275088080233225279830302955]
[12, 3.0903275680804514532022819491404]
[13, 3.0903275757136553115380751448278]
[14, 3.0903275766617688253756194898005]
[15, 3.0903275767756406402260131593259]
[16, 3.090327576788894242407995770567]
[17, 3.0903275767903920462396714344038]
[18, 3.0903275767905566816173514994191]
[19, 3.0903275767905743095256530578222]
[20, 3.0903275767905761506405993119406]
[21, 3.0903275767905763384415503552701]
[22, 3.0903275767905763571714370744]
[23, 3.0903275767905763589996761785457]
[24, 3.0903275767905763591744970397934]
[25, 3.0903275767905763591908872505933]
[26, 3.0903275767905763591923950587593]
[27, 3.0903275767905763591925312631695]
[28, 3.0903275767905763591925433527137]
[29, 3.0903275767905763591925444077608]
[30, 3.0903275767905763591925444983399]
[31, 3.0903275767905763591925445059943]
[32, 3.0903275767905763591925445066313]
[33, 3.0903275767905763591925445066835]
[34, 3.0903275767905763591925445066878]
[35, 3.0903275767905763591925445066881]
[36, 3.0903275767905763591925445066881]
[37, 3.0903275767905763591925445066881]
>> 

Wolfram Alpha identifies this as the square of C(NR), which turns out to be the square of something called the nested radical constant.  So I took the square root:

>> sqrt(S)
ans =
1.7579327566180045327088196382181

and find

sqrt(1+sqrt(2+sqrt(3)+ ...))

where the sqrt is taken one step further, explaining the square in what what Wolfram Alpha says.

That constant's digits are laid out in OEIS's A072449.

So the number basically itself defines a fundamental constant, as Wolfram Mathworld says that "No closed-form expression is known for this constant", referring to 1.7579..., and by extension to that number squared, which is the current solution.

  Posted by Charlie on 2023-04-12 08:01:50
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