why is $ 2 = \frac{5}{1+\frac{8}{4+\frac{11}{7 + \frac{14}{10 + \dots}}} } $

Question

Why is $ 2 = \cfrac{5}{1+\cfrac{8}{4+\cfrac{11}{7 + \cfrac{14}{10 + \ddots}}} } $

where the sequences $5,8,11,14,\dots$ and $1,4,7,10,\dots$ are of the form $5 + 3 n$ and $1 + 3n$.

(This converges on both even and uneven iterates)

I was surprised this is an integer.

Maybe it would help to rewrite this generalized continued fraction into a "normal" simple continued fraction. But I believe that would give us coefficients that generalized the double factorial to a sort of " triple factorial " meaning $ f(n) = n \cdot f(n-3) \cdot f(n-6) \cdot f(n-9) \cdots $ and I have almost no skills or understanding of those.

Maybe some transformation formula's make this easy, but Im not seeing it.

Answer 1

A relative simple solution can be found based on a fews facts about continued fractions which I will present briefly.

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Brief summary of (positive) general continued fractions: For any sequences of positive numbers $(a_n)$ and $(b_n)$ defined the maps $$M_n(x)=\tfrac{a_1}{b_1+\tfrac{a_2}{b_2+\ddots \begin{matrix} \\ +\tfrac{a_{n-1}}{b_{n-1}+\tfrac{a_n}{b_n+x}} \end{matrix}}}$$ The terms $M_n(0)$ are known as converts of the continued fraction $$\tfrac{a_1}{b_1+\tfrac{a_2}{b_2+\ddots \begin{matrix} \\ +\tfrac{a_{n-1}}{b_{n-1}+\ddots} \end{matrix}}}$$ It is easy to check that
$$M_n(x)=M_{n-1}\Big(\frac{a_n}{b_n+x}\Big)$$ By induction it follows that $$M_n(x)=\frac{p_{n-1}x+p_n}{q_{n-1}x+q_n}$$ where \begin{align} p_n&=a_np_{n-2}+ b_np_{n-1},\qquad p_0=0,\, p_1=a_1\\ q_n&=a_nq_{n-2}+b_nq_{n-1},\qquad q_0=1,\, q_1=b_1 \end{align} and $$\begin{align} p_{n-1}q_n-q_{n-1}p_n=(-1)^na_1\cdot\ldots \cdot a_n\tag{0}\label{zero} \end{align}$$ Hence $$\begin{align} \frac{p_{n-1}}{q_{n-1}}-\frac{p_n}{q_n}=(-1)^n\frac{a_1\cdot\ldots\cdot a_n}{q_{n-1}q_n}\tag{1}\label{one} \end{align} $$ and $$\begin{align} \frac{p_n}{q_n}-\frac{p_{n-2}}{q_{n-2}}=(-1)^{n-1}b_n\frac{a_1\cdot\ldots\cdot a_{n-1}}{q_nq_{n-2}}\tag{2}\label{two} \end{align}$$ More generally $$\begin{align} M_n(x)-M_n(y)=(-1)^n(x-y)\frac{a_1\cdot\ldots\cdot a_n}{\big(q_{n-1}x+q_n\big)\big(q_{n-1}y+q_n\big)}\tag{3}\label{three} \end{align}$$ Identities \eqref{one} and \eqref{two} imply that $$\frac{p_{2n}}{q_{2n}}<\frac{p_{2n+2}}{q_{2n+2}}<\frac{p_{2m+1}}{q_{2m+1}}<\frac{p_{2m-1}}{q_{2m-1}}$$ for all $m, n$. Thus, there are numbers $0<\alpha\leq \beta$ such that $p_{2n}/q_{2n}\xrightarrow{n\rightarrow \infty}\alpha$ and $p_{2n+1}/q_{2n+1}\xrightarrow{n\rightarrow\infty}\beta$.

The following result gives sufficient conditions for convergence of the convergents $M_n(0)=\frac{p_n}{q_n}$.

Lemma: If $\liminf_n\frac{b_{n-1}b_n}{a_n}>0$, then $\frac{p_n}{q_n}$ converges.

A proof of this result (with limit instead of $\liminf$) can be found in the classic high shool textbook Hall, H.S. and Knight, S. R., Higher algebra, 4th edition, 1889. pp. 362. A short proof of the Lemma is also shown at the end of this posting.

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Solution t the OP: For the problem in the OP consider $a_n=3n+2$ and $b_n=3n-2$ for each $n\in\mathbb{N}$. The key part of the whole solution is the observation made by Ivan Kaznacheyeu in his comment. Notice that $M_1(x)=\frac{5}{1+x}$, and that $2=M_1(1+\frac12)$. Let $f(n)=1+\frac{1}{n+1}$. This sequence satisfies $$ f(n-1)=\frac{3n+2}{(3n-2)+f(n)}$$ From this, $$ M_n(f(n))=M_{n-1}\big(\frac{3n+2}{(3n-2)+f(n)}\big)=M_{n-1}(f(n-1))$$ Thus, if $M_{n-1}(f(n-1))=2$, we have that $M_n(f(n))=2$. This was Ivan Kaznacheyeu's insight into the problem. The rest of the solution, as we will see is routine.

Since $$\frac{b_{n-1} b_n}{a_{n-1}}=\frac{(3n-5)(3n-2)}{3n-1}\xrightarrow{n\rightarrow\infty}\infty,$$ The sequence $M_n(0)=p_n/q_n$ converges.

As Ivan Kaznacheyeu mentioned in his comment, for any $n\in\mathbb{N}$ $$2=M_n(f(n)),\qquad\text{where}\quad f(n)=1+\frac{1}{n+1}$$ Consequently, from \eqref{three} $$\begin{align} 2-\frac{p_n}{q_n}=M_n(f(n))-M_n(0)=(-1)^nf(n)\frac{a_1\cdot\ldots\cdot a_n}{\big(q_{n-1}f(n)+q_n\big)q_n}\tag{4}\label{four}\end{align} $$ Identity \eqref{four} yields $$\frac{p_{2n}}{q_{2n}}<2<\frac{p_{2m+1}}{q_{2m+1}},\qquad \forall \,n,m\in\mathbb{N} $$ Hence, $\frac{p_n}{q_n}$ converges to $2$.

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Proof of Lemma: From \eqref{zero} $$\frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}}=-\frac{a_nq_{n-2}}{q_n}\Big(\frac{p_{n-1}}{q_{n-1}}-\frac{p_{n-2}}{q_{n-2}}\Big)$$ Then $$ \frac{a_nq_{n-2}}{q_n}=\frac{a_nq_{n-2}}{a_nq_{n-2} + b_nq_{n-1}}=\frac{1}{1+\frac{b_nq_{n-1}}{a_nq_{n-2}}} $$ and further $$\frac{b_nq_{n-1}}{a_nq_{n-2}} = \frac{b_n(a_{n-1}q_{n-3}+b_{n-1}q_{n-2})}{a_nq_{n-2}}=\frac{b_na_{n-1}q_{n-3}}{a_nq_{n-2}}+\frac{b_nb_{n-1}}{a_n} $$ As the term $\frac{b_na_{n-1}q_{n-3}}{a_nq_{n-2}}>0$ for all $n$, the assumption in the Lemma implies that for some $c>0$ $$0<\frac{a_nq_{n-2}}{q_n}<\frac{1}{1+c}$$ for all $n$ large enough. From this, it follows that the difference $\Big|\frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}}\Big|\xrightarrow{n\rightarrow\infty}0$, that is $\alpha=\beta$.

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Answer 2

Too long for a comment:
It appears that by cutting of the head at the top, one gets: $1+1,1+1/2,1+1/3,1+1/4,1+1/5,...,1+1/r$

(*Mathematica start*)
r = 20;
n = Table[5 + 3*(n - 1), {n, r, 100}];
d = Table[1 + 3*(n - 1), {n, r, 100}];
f[y_, {m_, d_}] := m/(d + y);
continuedFraction = 
 Fold[f, Last@n/Last@d, Reverse@Most@Transpose@{n, d}]
"The output appears to be equal to r"
(N[continuedFraction, 20] - 1)^-1
(*end*)

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$r=1$ gives: continuedFraction $= 2$
for sequences $n$ and $n-2$

(*Mathematica start*)
r = 10;
n = Table[n, {n, r, 100}]
d = Table[(n - 2), {n, r, 100}]
f[y_, {m_, d_}] := m/(d + y);
continuedFraction = 
 Fold[f, Last@n/Last@d, Reverse@Most@Transpose@{n, d}]
N[continuedFraction, 20]
"The output appears to be equal to r"
(N[continuedFraction, 20] - 1)^-1
(*end*)

Answer 3

Assume the RHS of the equation as $A_n$ so we have the following recurrence relation: $$ A_n = \frac{3n + 2}{3n - 2 + A_{n + 1}}. $$ Claim: $$A_n = \frac{n + 1}{n}.$$ Proof:

Step of the induction: $$A_{n + 1} = \frac{n + 2}{n + 1} \Rightarrow A_n = \frac{3n + 2}{3n - 2 + \frac{n + 2}{n + 1}} = \frac{n + 1}{n}.$$

Base of the induction:

Limit of the $A_n$ in infinity according to our closed form solution: $$A_{\infty} = \lim_{n \to \infty}{\frac{n + 1}{n}} = 1.$$

Limit of the $A_n$ in infinity according to the given recurrence relation: $$ x = \lim_{n \to \infty}{A_n} = \lim_{n \to \infty}{\frac{3n + 2}{3n - 2 + A_{n + 1}}} = \lim_{n \to \infty}{\frac{3n + 2}{3n - 2 + x}} = 1.$$

So the base of the induction is correct as well.

So $A_n = \frac{n + 1}{n}$.

Your question asks about the value of $A_{n = 1}$ and we have $ A_{1} = \frac{n + 1}{n} = \frac{1 + 1}{1} = 2$

Answer 4

Tommy1729 aka my mentor Tom Raes wrote to me :

Proof (sketch)

We want to prove cutting of the head at the top gives the result $1+1/n$ from which the case ' equals $2$ ' follows.

Now if we get close to $1+1/n$ then it must really be that ($1+ 1/n$) because the starting value for continued fractions is irrelevant just like $\dfrac{1}{1+\dfrac{1}{1+\dfrac{...}{1+x}}}$ converges to the golden mean independant of any fixed positive $x$.

Lets take $n$ and $k$ positive integers.

$ \dfrac{5 + 3(n-1)(1 + e_1)}{1+ 1/n} - [1 + 3(n-1)] = 1 + 1/(n+1) + e_1[\dfrac{5 + 3(n-1)}{1 + 1/n}]$

going down in the continued fraction $k$ times clearly gives us

$e_k$ follows $ (1+e_k)^{-1} (1 + 1/(n+k-1)) = 1 + 1/(n+k-1) + e_{k-1}[\frac{5 + 3(n+k-3)}{1 + 1/(n+k-2)}]$

From then we can conclude that for sufficienly large $n+k = v$ we get in the limit

$(1 + e_v)^{(-1)} = "1" + e_{v-1} * "\infty"$.

where " means approximating.

Notice $e_v$ going towards (positive or negative) $\infty $ cannot keep on satisfying the equations. So that never happens. Similar with $e_v$ not converging to $0$ nor $-1$; that cannot keep on satisfying the equations.

If $e_v$ converges to $-1$ we get in the limit

$(1+ 1/v)^{-1} (1 + e_v) = 0/1 = 0$

so

$ (1 + 1/v + w)^{-1} = 0$

Therefore the starting value converges to infinity because $w$ does.

But that is not valid.

( compare with 3 = $\dfrac{1}{1+\dfrac{1}{1+\dfrac{...}{1+x_n}}}$ then $x_n$ diverges to infinity because the value 3 should be the golden mean. )

So the only solutions $1 + e_v$ that converge are being $1$ or converging to $1$ ( at rate $O ( 1/v^3)$ ).

SO we get closer to $1+1/n$ as desired. And thus it actually IS $1+1/n$.

QED

Ps: the other continued fractions I sent you are harder to prove.