I don’t have nice pictures like Adi but I might as well include my solutions from above.

(1) Call the value of that expression x. x=sqrt(1+sqrt(1+…)). Square both sides. Then, subtract 1 from both sides. Since the expression continues infinitely, notice now that x^2-1 is the same as x. Solve using the quadratic formula (take the positive root).

(2) I propose that sqrt(1+nsqrt(1+(n+1)sqrt(…))) is in fact just n+1. Take n=0. All of the inner terms fall away and you’re left with sqrt(1)=1=n+1.

Assume it’s true for n. n+1=sqrt(1+nsqrt(1+(n+1)sqrt(…))). Square both sides. Then, subtract 1 from both sides. Divide both sides by n. You’ve got ((n+1)^2-1)/n=sqrt(1+(n+1)sqrt(1+(n+2)sqrt(…))). RHS is the expression for n+1. LHS is (n^2+2n+1-1)/n=n+2. Formula is true for n+1. By induction, blah blah blah.

We have x(0) = 1 and we know that: x(n)^2 = 1 + nx(n+1).

If x(n) = n + 1, then: (n+1)^2 = 1 + nx(n+1), so: x(n+1) = n+2, QED.

Results: x(2) = 3.

How can we arrive to the hypothesis x(n) = n+1?

We note that: x(2)^2 + x(3)^2 = (x(3) + 1)^2, so chances are that (x(2), x(3), x(3)+1) could be a Pythagorean triple (the most famous one: 3^2 + 4^2 = 5^2). If x(2) would be 3 it’s easy to observe that also x(3) would be 4, x(4) = 5, etc, so we can make the hypothesis x(n) = n+1

Haha, you caught me there. I stayed up way too late tonight until I was able to paper over the trouble spot with some complex analysis. Is there are a solution using purely elementary methods?

No suggestions on putting together an elementary method solution? If the puzzle is an early one by Ramanujan, I doubt the original solution was done with complex analysis.

I did write a recursive function, but analysis is needed to prove that the sequence converges.

Define r_n(z)=sqrt(1+nz). Then, a finite approximation to some degree n is s(n,z)=r_2r_3…*r_n(z). The * there is function composition. The answer is the limit n->inf of s(n,1).

Working this morning I could replace the argument with a real-valued one, so the z can become an x.

I don’t see a purely algebraic means of doing this without a step that handwaves over the part "and then repeat this process forever".

I’m happy with my solution using analysis so I’m not going to push this any further. I would like to see a more classical technique if one exists. But, this was done at the very beginning of the age of rigor so perhaps handwaving was good enough at the time given that the result is indeed correct.

OK – I am going now to give one more hint to put in evidence a fundamental problem with all the approaches above.

Your formulas are correct. In fact you can write directly:

3 = sqrt(1+24)

= sqrt(1+2sqrt(1+35))

= sqrt(1+2sqrt(1+3sqrt(1+46)))

= sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+57))))

…

= sqrt(1+2sqrt(1+… sqrt(1+(n-1)(n+1))…))

… (by taking to the limit)

= sqrt(1+2sqrt(1+3sqrt(1+4…)))

But the trouble is not the convergence. The trouble is that "3" is not the only number with this property. In fact, it is easy to see that for an infinity of real numbers you can build a series which ends up with exactly the same infinite radical expression like the one above.

For example, let’s pick up:

X_1 = 4

X_2 = 1/2 * (X_1 ^ 2 – 1)

X_3 = 1/3 * (X_2 ^ 2 – 1)

…

X_n = 1/n * (X_(n-1) ^ 2 – 1)

…

It is easy to see that X_i is well-defined and monotonically growing. Then, you can write:

X_1 = sqrt(1+2X_2)

= sqrt(1+2sqrt(1+3X_3))

= sqrt(1+2sqrt(1+3sqrt(1+4X_4)))

= sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+5X_5))))

…

= sqrt(1+2sqrt(1+… sqrt(1+(n-1)X_(n-1))…))

… (by taking to the limit) …

= sqrt(1+2sqrt(1+3sqrt(1+4…)))

Therefore:

4 = sqrt(1+2sqrt(1+3sqrt(1+4*…)))

Huh?!

So it’s not enough to proof convergence if you start with "3". You also need to _uniquely_ define what that infinite radical expression means.

The first is (1+sqrt(5))/2 by simple rearrangement and the quadratic formula.

The second is 3 by induction (which reveals a whole family of nice formulas).

First one, solve for x=sqrt(1+x)…

This gives (1+sqrt(5))/2 which I believe is the golden ratio.

I don’t have nice pictures like Adi but I might as well include my solutions from above.

(1) Call the value of that expression x. x=sqrt(1+sqrt(1+…)). Square both sides. Then, subtract 1 from both sides. Since the expression continues infinitely, notice now that x^2-1 is the same as x. Solve using the quadratic formula (take the positive root).

(2) I propose that sqrt(1+n

sqrt(1+(n+1)sqrt(…))) is in fact just n+1. Take n=0. All of the inner terms fall away and you’re left with sqrt(1)=1=n+1.Assume it’s true for n. n+1=sqrt(1+n

sqrt(1+(n+1)sqrt(…))). Square both sides. Then, subtract 1 from both sides. Divide both sides by n. You’ve got ((n+1)^2-1)/n=sqrt(1+(n+1)sqrt(1+(n+2)sqrt(…))). RHS is the expression for n+1. LHS is (n^2+2n+1-1)/n=n+2. Formula is true for n+1. By induction, blah blah blah.Nicolas:

Interesting, but can you detail how do you go in your induction step from n=0 to n=1?

2.

Let be: x(n) = sqrt(1 +n

sqrt(1 +(n+1)sqrt(1+…We have to calculate x(2).

By induction we will prove that: x(n) = n +1.

We have x(0) = 1 and we know that: x(n)^2 = 1 + n

x(n+1).

x(n+1), so: x(n+1) = n+2, QED.If x(n) = n + 1, then: (n+1)^2 = 1 + n

Results: x(2) = 3.

How can we arrive to the hypothesis x(n) = n+1?

We note that: x(2)^2 + x(3)^2 = (x(3) + 1)^2, so chances are that (x(2), x(3), x(3)+1) could be a Pythagorean triple (the most famous one: 3^2 + 4^2 = 5^2). If x(2) would be 3 it’s easy to observe that also x(3) would be 4, x(4) = 5, etc, so we can make the hypothesis x(n) = n+1

Haha, you caught me there. I stayed up way too late tonight until I was able to paper over the trouble spot with some complex analysis. Is there are a solution using purely elementary methods?

Smells like Ramanujan spirit ðŸ™‚

(n + 1)^2 = 1 + n

((n + 1) + 1)

sqrt(1 + (n+1)n + 1 = sqrt(1 + n

sqrt(1 + (n+2)sqrt(…Several answers to my previous puzzle&nbsp;reminded me about an old result:

Theorem: All horses have…

No suggestions on putting together an elementary method solution? If the puzzle is an early one by Ramanujan, I doubt the original solution was done with complex analysis.

No complex analysis should be required, AFAIK. Hint: try to express the problem recursively.

(But an "induction" would not be useful here)

I did write a recursive function, but analysis is needed to prove that the sequence converges.

Define r_n(z)=sqrt(1+nz). Then, a finite approximation to some degree n is s(n,z)=r_2

r_3…*r_n(z). The * there is function composition. The answer is the limit n->inf of s(n,1).Working this morning I could replace the argument with a real-valued one, so the z can become an x.

I don’t see a purely algebraic means of doing this without a step that handwaves over the part "and then repeat this process forever".

I’m happy with my solution using analysis so I’m not going to push this any further. I would like to see a more classical technique if one exists. But, this was done at the very beginning of the age of rigor so perhaps handwaving was good enough at the time given that the result is indeed correct.

In essence, your solution is very close to the final proof – the only missing part is showing why "3" is the correct answer and not, say, "3.1".

I will post a complete proof soon.

n+p+2 = sqrt(1 + (n+p+1)

(n+p+3))

sqrt(1 + (n+p+1)n+p+1 = sqrt(1 + (n+p)

(n+p+3)))

sqrt(1 + (n+p)n+p = sqrt(1 + (n+p-1)

sqrt(1 + (n+p+1)(n+p+3))))…

n+1 = sqrt(1 + n

sqrt(1 + (n+1)sqrt(1 +…(n+p)sqrt(1 + (n+p+1)(n+p+3) = x(n, p)we have to prove the convergence of the x(n, p) when p->inf

But x(n, p) = x(n, p+1) = n + 1 because n+p+3 = sqrt(1+(n+p+2)*(n+p+4))

ðŸ™‚

OK – I am going now to give one more hint to put in evidence a fundamental problem with all the approaches above.

Your formulas are correct. In fact you can write directly:

3 = sqrt(1+2

4)

sqrt(1+3= sqrt(1+2

5))

sqrt(1+3= sqrt(1+2

sqrt(1+46)))= sqrt(1+2

sqrt(1+3sqrt(1+4sqrt(1+57))))…

= sqrt(1+2

sqrt(1+… sqrt(1+(n-1)(n+1))…))… (by taking to the limit)

= sqrt(1+2

sqrt(1+3sqrt(1+4…)))But the trouble is not the convergence. The trouble is that "3" is not the only number with this property. In fact, it is easy to see that for an infinity of real numbers you can build a series which ends up with exactly the same infinite radical expression like the one above.

For example, let’s pick up:

X_1 = 4

X_2 = 1/2 * (X_1 ^ 2 – 1)

X_3 = 1/3 * (X_2 ^ 2 – 1)

…

X_n = 1/n * (X_(n-1) ^ 2 – 1)

…

It is easy to see that X_i is well-defined and monotonically growing. Then, you can write:

X_1 = sqrt(1+2X_2)= sqrt(1+2

sqrt(1+3X_3))= sqrt(1+2

sqrt(1+3sqrt(1+4X_4)))

sqrt(1+3= sqrt(1+2

sqrt(1+4sqrt(1+5X_5))))

sqrt(1+… sqrt(1+(n-1)…

= sqrt(1+2

X_(n-1))…))

sqrt(1+3… (by taking to the limit) …

= sqrt(1+2

sqrt(1+4…)))Therefore:

4 = sqrt(1+2

sqrt(1+3sqrt(1+4*…)))Huh?!

So it’s not enough to proof convergence if you start with "3". You also need to _uniquely_ define what that infinite radical expression means.

Once you have the sequence above, it’s tedious but not impossible to prove that it converges to 3.

From the definition, s(n+1,z)=s(n,sqrt(1+(n+1)z)).

You can prove by induction that s(n,n+2)=3.

s’>0 and s”<0 (more side proofs) with respect to z so s is monotonic and concave.

So, the linear interpolate is less than the actual point.

s(n,b)>s(n,c)-(s(n,c)-s(n,a))

(c-b)/(c-a)Pick a=1,b=sqrt(1+(x+1)),c=x+2,n=x.

s(x,sqrt(1+(x+1)))>s(x,x+2)-(s(x,x+2)-s(x,1)).

(x+2-b)/(x+1).s(x+1,1)>3-(3-s(x,1))

3-s(x+1,1)<(x+2-b)/(x+1)

(3-s(x,1)).

(3-s(x,1)).3-s(x+1,1)<(x+2-b)/x

Now we’re in a position to examine this as lim x->inf.

When x is large we get

3-s(x+1,1)<(x-1)/x

(3-s(x,1)).Take lim x->inf of both sides. Look at the RHS first.

(x-1)/x(3-s(x,1))=(1-1/x)(3-s(x,1)).Apply the linear interpolation formula above during the limit.

(1-1/x)(1-1/(x-1))(1-1/(x-2))…(3-s(C,1))going down to some constant C when starting from finite x.

Take the log of both sides so

ln(3-s(x+1,1)) <

(1-1/(x-1))ln((1-1/x)

(1-1/(x-2))…*(3-s(C,1))) <ln(1-1/x)+ln(1-1/(x-1))+ln(1-1/(x-2))+…

Apply ln(1+q)<=q giving

-1/x-1/(x-1)-1/(x-2)-… all as x->inf.

That’s the harmonic sequence so the limit diverges to -inf.

So, lim x->inf ln(3-s(x+1,1))=-inf.

lim x->inf 3-s(x+1,1)=0.

lim x->inf s(x,1)=3.

Very nice proof! I couldn’t find any mistake so far.

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