Let $x_{n} = \sqrt{1 +\sqrt{2 + \sqrt{3 + \dots \sqrt{n}}}}$. Show $\lim_{n \rightarrow \infty} x_{n}$ exists.

Question

Let $x_{n} = \sqrt{1 +\sqrt{2 + \sqrt{3 + \dots \sqrt{n}}}}$. show $\lim_{n \rightarrow \infty} x_{n}$ exists.

To do this the problem has been broken down into three pieces:

a) Show that $x_{n} < x_{n+1}$ (I've completed this)

b) Establish $(x_{n + 1})^2 < 1 + \sqrt{2} x_{n}$ Hint: square $x_{n+1}$ and factor a 2 out of the square root. (Having trouble with this)

c) Conclude that $x_{n} \leq 2$ and thus $\lim_{n \rightarrow \infty} x_{n}$ exists. (Having trouble with this)

Attempts

Starting with part b):

After following the hint and doing some algebra the problem has been reduced to trying to establish that:

$$\sqrt{3 + \dots \sqrt{n + \sqrt{n+1}}} < 2 \cdot \sqrt{2 + \sqrt{3 + \dots \sqrt{n}}}$$

At this point we were given a "fact" that $\sqrt{2n} \leq n-1$

Here is where the confusion lies for me. I know I'm supposed to get a chain of inequalities, but I can't seem to breakdown the expression into the necessary parts. Some help here would be nice.

Part c):

I am stumped. Mainly because I'm not sure how to show that this is bounded above. I get the implication that once we establish that this sequence is bounded above, then by the monotone convergence theorem we can establish the sequence converges. My issue here is that at least on the surface taking the limit of the sequence $x_{n}$ it would seem that the value would extend beyond 2 because I am summing up an infinite amount of objects. So....How should I proceed?

Answer 1

(a) is clear because of $\sqrt n = \sqrt n+0<\sqrt{n+\sqrt{n+1}}$ and compare the long iterated radicals.

(b) is also clear because of $$ \begin{aligned} 1+\sqrt 2\cdot x_{n} &= 1+2 \sqrt{1 +\sqrt{2 + \sqrt{3 + \dots \sqrt{n}}}} \\ &= 1+ \sqrt{2^2 +2^2\sqrt{2 + \sqrt{3 + \dots \sqrt{n}}}} \\ &= 1+ \sqrt{2^2 +\sqrt{2\cdot 2^4 + 2^4\sqrt{3 + \dots \sqrt{n}}}} \\ &= 1+ \sqrt{2^2 +\sqrt{2\cdot 2^4 + \sqrt{3\cdot 2^8 + 2^8\dots \sqrt{n}}}} \\ &\qquad\text{ and push the powers of two till the end,} \\ &\qquad\text{ getting instead of the final / deepest $n$ in $1+\sqrt2\cdot x_n$} \\ &\qquad\text{ something $>$ then the final / deepest $(n+1)$ in $x_{n+1}^2$} \\ &> x_{n+1}^2\ , \end{aligned} $$ (Later EDIT: Thanks to the observation of dc3rd, note that in both $1+\sqrt 2\cdot x_n$ and $x_{n+1}^2$ we have the same number of iterated radicals, so we compare them one by one from the deepest one to the outer one.)

(c) is finally simple by induction, since $x_1=1\le 2$, and assuming inductively $x_n\le 2$ we get $x_{n+1}<\sqrt{1+\sqrt 2\cdot x_n} \le\sqrt{1+\sqrt 2\cdot 2} \le\sqrt{1+3} =2 $.

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Later edit: Here are some values of the sequence, numerically calculated in pari/gp to some modest precision:

? \p 50
? {x(N) = a=0.; for(k=1, N, a=sqrt(N+1 -k +a)); a;}
? values = [1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70];
? for(j=1, 16, k=values[j]; print("x(", k, ") = ", x(k)))
x(1) = 1.0000000000000000000000000000000000000000000000000
x(2) = 1.5537739740300373073441589530631469481645834994103
x(3) = 1.7122650649295326242302679779342230870015699173989
x(4) = 1.7487627132551437866964866764318493705517606065467
x(5) = 1.7562384875823431864643909051164497734275724334927
x(6) = 1.7576412350415822329154094314563757401237452707488
x(7) = 1.7578856460964371138824575295846204229591662122319
x(8) = 1.7579255575682606866360226374254570646166465189101
x(9) = 1.7579317105145658014667625802995697218663165224376
x(10) = 1.7579326113938309894205758132194406438710917961082
x(20) = 1.7579327566180044733914355091545785526727096291796
x(30) = 1.7579327566180045327088196358436875517051703277110
x(40) = 1.7579327566180045327088196382181385276330617489975
x(50) = 1.7579327566180045327088196382181385276531999221468
x(60) = 1.7579327566180045327088196382181385276531999221468
x(70) = 1.7579327566180045327088196382181385276531999221468