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I would like to investigate the convergence of

$$sqrt{1+sqrt{2+sqrt{3+sqrt{4+sqrtldots}}}}$$

Or more precisely, let $$begin{align}
a_1 & = sqrt 1\
a_2 & = sqrt{1+sqrt2}\
a_3 & = sqrt{1+sqrt{2+sqrt 3}}\
a_4 & = sqrt{1+sqrt{2+sqrt{3+sqrt 4}}}\
&vdots
end{align}$$

Easy computer calculations suggest that this sequence converges rapidly to the value 1.75793275661800453265, so I handed this number to the all-seeing Google, which produced:

• OEIS A072449


• "Nested Radical Constant" from MathWorld

Henceforth let us write $sqrt{r_1 + sqrt{r_2 + sqrt{cdots + sqrt{r_n}}}}$ as $[r_1, r_2, ldots r_n]$ for short, in the manner of continued fractions.

Obviously we have $$a_n= [1,2,ldots n] le underbrace{[n, n,ldots, n]}_n$$

but as the right-hand side grows without bound (It's $O(sqrt n)$) this is unhelpful. I thought maybe to do something like:

$$a_{n^2}le [1, underbrace{4, 4, 4}_3, underbrace{9, 9, 9, 9, 9}_5, ldots,
underbrace{n^2,n^2,ldots,n^2}_{2n-1}] $$

but I haven't been able to make it work.

I would like a proof that the limit $$lim_{ntoinfty} a_n$$
exists. The methods I know are not getting me anywhere.

I originally planned to ask "and what the limit is", but OEIS says "No closed-form expression is known for this constant".

The references it cites are unavailable to me at present.

For any $nge4$, we have $sqrt{2n} le n-1$. Therefore
begin{align*}
a_n
&le sqrt{1+sqrt{2+sqrt{ldots+sqrt{(n-2)+sqrt{(n-1) + sqrt{2n}}}}}}\
&le sqrt{1+sqrt{2+sqrt{ldots+sqrt{(n-2)+sqrt{2(n-1)}}}}}\
&leldots\
&le sqrt{1+sqrt{2+sqrt{3+sqrt{2(4)}}}}.
end{align*}
Hence ${a_n}$ is a monotonic increasing sequence that is bounded above.

The first number $1$ is a nuisance, so at first we disregard it.

We proceed by induction, and deal with finite nested radicals that start with $sqrt{k+sqrt{(k+1)+cdots}}$, where $kge 2$.

We will show that such a radical is $lt 2k$, by induction on depth. The result is certainly true for all nested radicals of depth $1$, since $sqrt{q}lt 2q$.

For the induction step, a nested radical of depth $n$ that starts with $k$ is $sqrt{k+R}$, where $R$ is a nested radical of depth $n-1$ that starts with $k+1$. By the induction assumption, we have $Rlt 2k+2$. But then $sqrt{k+R}lt sqrt{3k+2}lt 2k$ if $k ge 2$.

So (finite) nested radicals of any depth that start with $2$ are $lt 4$. The sequence of nested radicals is clearly increasing, so it converges. It follows that the nested radical of the post is $le sqrt{1+4}$.

An easy sloppy way to see it:
You know that $sqrt{1+sqrt{1+sqrt{1+sqrt{1+dots}}}}=phi$. Now multiply by $2$ to get $2phi=2sqrt{1+sqrt{1+sqrt{1+sqrt{1+dots}}}}=sqrt{4+4sqrt{1+sqrt{1+sqrt{1+dots}}}}=sqrt{4+sqrt{16+sqrt{256+sqrt{256^2+dots}}}}>sqrt{1+sqrt{2+sqrt{3+sqrt{4+dots}}}}$. Of course this should be done more rigorous.

If $( a_k )_{kinmathbb{N}}$ is any sequences of positive numbers such that:

$$0 le a_k le alpha lambda^{2^k}quadtext{ for some }quad alpha, lambda in mathbb{R}_{+}$$

Using same convention $;[r_1,r_2ldots] = sqrt{r_1 + sqrt{r_2 + ldots}};$ as in the question,
we have:

$$begin{align}
[a_n] & le sqrt{alpha lambda^{2^n}} = sqrt{alpha}lambda^{2^{n-1}} = [alpha] lambda^{2^{n-1}}\
implies [a_{n-1},a_n ] &le sqrt{alphalambda^{2^{n-1}}+sqrt{alpha}lambda^{2^{n-1}}}
=sqrt{alpha+sqrt{alpha}}lambda^{2^{n-2}} = [alpha,alpha]lambda^{2^{n-2}}\
implies [a_{n-2},a_{n-1},a_n]&le sqrt{alphalambda^{2^{n-2}} + sqrt{alpha+sqrt{alpha}}lambda^{2^{n-2}}} = [alpha,alpha,alpha]lambda^{2^{n-3}}\
&;vdots\
implies [a_1,ldots,a_n] & le underbrace{ [ alpha,ldots,alpha ] }_{ntext{ terms}} lambda\
implies [a_1, ldots,a_n] & le [ alpha, alpha, ldots ]lambda = frac{1 + sqrt{1+4alpha}}{2}lambda
end{align}$$

Since $n le sqrt{2}^{2^n-2}$, we can take $alpha = frac12$ and $lambda = sqrt{2}$ to get:

$$[1,2,ldots,n] le underbrace{ [ frac12,ldots,frac12 ]}_{ntext{ terms}} lambda le frac{1+sqrt{3}}{sqrt{2}} sim 1.931851 $$

To get a better bound, observe for any $m, k in mathbb{Z}_{+}$, we have:

$$ m + k - 1 le frac{m^2}{m+1}left(sqrt{frac{m+1}{m}}right)^{2^k}$$

Using the same approach as above, we get:
$$[m,m+1,m+2,ldots] le frac{sqrt{m+1}+sqrt{4m^2+m+1}}{2sqrt{m}}$$

Take $m = 3$, we already get a bound accurate up to $O(10^{-2})$.

$$[3,4,ldots] le frac{1+sqrt{10}}{sqrt{3}}
implies [1,2,3,ldots] le sqrt{1+sqrt{2+frac{1+sqrt{10}}{sqrt{3}}}} sim 1.760214368
$$

Take a positive sequence ${a_n}$ and a constant $c>0$ such that $sqrt{a_{n+1}}<ca_n$.

Set $b_n=sqrt{a_1+sqrt{a_2+cdots sqrt{a_n}}}$. By induction, $$log_{c+1} left(frac{b_n}{sqrt{a_1}}right)<sum_{i=1}^{n-1}2^{-i}<1$$
So $b_n<(c+1)sqrt{a_1}$ and $b_n$ is monotonic increasing; by the Monotone Convergence Theorem, $lim_{ntoinfty}b_n$ exists. Taking $a_n=n$ and $c>sqrt{2}$ answers this question.

By linear approximations.

Since the sequence is increasing, it converges iff it doesn't go to infinity. Therefore we need only to construct an upper bound.

Notice that all tangents to $sqrt{x}$ are always above $sqrt{x}$. The slope of each tangent is $frac12frac1{sqrt x}$; but we can construct a simpler bound by noticing that the line through $(x,sqrt x)$ with slope $frac 1 2$ is above $sqrt{u}$ for all $u > x$ if $x geq 1$.

So $$sqrt{1 + sqrt{2 + sqrt{3 + dotsc}}} leq 1 + frac12(sqrt 2 + frac12(sqrt{3} + frac12(dotsc))) = sum_{i=1}^infty frac{sqrt i}{2^{i-1}}$$ and it's easier to see that that converges.