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We denote for integers $n\geq 1$ the $nth$ prime number as $p_n$. And for integers $k\geq 1$, I consider this recurrence relation $$a_{k+1}=\frac{a_k}{p_{k+1}}+\frac{1}{a_k},\tag{1}$$ with $a_1$ defined to be equal to $1$.

I've calculated some (few) terms of this sequence $a_k$. See here the first examples.

Examples of computations of some terms of our sequence $ \left\{ a_n\right\}_{n=1} ^\infty$:

1) Since $a_1=1$ then $a_2=\frac{a_1}{p_2}+(a_1)^{-1}=\frac{1}{3}+\frac{1}{1}=\frac{4}{3}$.

2) Since the third prime number is $p_3=5$ one has $$a_3=\frac{a_2}{p_3}+\frac{1}{a_2}=\frac{4/3}{5}+\frac{3}{4}=\frac{61}{60}.$$ 3) Similarly $a_4=\frac{61}{7\cdot 60}+(60/61)^{-1}\approx 1.1288$. $\square$

Thus our sequence starts as $a_1=1,a_2\approx 1.3333, a_3\approx 1.0117, a_4\approx 1.1288$ and similarly we can calculate $a_5\approx 0.9664,a_6\approx 1.1091$ or $a_7\approx 0.9669$.

To create this problem I was inspired in a recurrence that $\sqrt{2}$ solves as you can see from this WIkipedia, and now I am curious about how to check that the sequence in $(1)$ is convergent.

Question. Please, can you prove that $\left\{ a_n\right\}_{n=1} ^\infty$ defined from $(1)$ is convergent? Many thanks.

I know a a main tool in the theory of prime numbers is the Prime Number Theorem: $p_n\sim n\log n$ as $n\to\infty$.

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    $\begingroup$ have you tried to look for conditons under which $a_{k+1}>2$? $\endgroup$ – dfnu Oct 1 '17 at 14:53
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    $\begingroup$ Sorry an error and the expression I could not edit of $\Delta_k = \frac{1}{a_k} - a_k\frac{p_{k+1}-1}{p_{k+1}}$ $\endgroup$ – dfnu Oct 1 '17 at 19:02
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    $\begingroup$ I was able to show that both $\{a_{2n}\}$ and $\{a_{2n+1}\}$ converge. Also numerical computation with first 100k terms suggests that the error $|a_n - 1|$ decays at least as fast as $\mathcal{O}(1/\log n)$. $\endgroup$ – Sangchul Lee Oct 2 '17 at 9:34
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    $\begingroup$ One could easily prove that if $a_k \leq 1+\epsilon$ for all $k \geq k_0$ then $a_k \geq 1-\epsilon$ for all $k \geq k_0$ and i proved that $a_k \leq 1.03$ for all $k \geq 20$ thus $0.97 \leq a_k \leq 1.03$ for all $k \geq 20$. $\endgroup$ – Ahmad Oct 2 '17 at 20:25
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    $\begingroup$ Many thanks feel free to add you contribution as an answer. It enriches the post @Ahmad $\endgroup$ – user243301 Oct 2 '17 at 22:42
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This partially answers the question by showing that both $(a_{2n})$ and $(a_{2n+1})$ converge. The proof may not be an easy path, and particularly the first section can be quite boring. You may focus on only color-boxed statements and skip the rest of the preliminary.


1. Preliminary

Definition. For $p \geq 2$ we define $f_p : (0, \infty) \to (0, \infty)$ by

$$f_p(x) = \frac{x}{p} + \frac{1}{x}$$

Also we set $g_{p,q} = f_q \circ f_p$ for $q \geq p \geq 2$.

We would like to investigate these functions. Then it is immediate that the equation $f_p'(x) = 0$ has a unique zero $x = \sqrt{p}$ on $(0, \infty)$. Using this, we can solve $g_{p, q}'(x) = f_q'(f_p(x)) f_p'(x) = 0$.

  • One zero comes from $f_p'(x) = 0$, yielding $x = \sqrt{p}$.

  • There are two other zeros that come from $f_q'(f_p(x)) = 0$ or equivalently $f_p(x) = \sqrt{q}$. Solving this, we obtain two zeros

    $$ \bbox[#fff9e3,border:1px solid #ffeb8e,12px]{ \alpha_{p,q} := \frac{p\sqrt{q} - \sqrt{ p (pq - 4)} }{2} } \quad \text{and} \quad \beta_{p,q} := \frac{p\sqrt{q} + \sqrt{ p (pq - 4)} }{2} \tag{1} $$

    As for the smaller zero, we have $\alpha_{p,q} = \frac{2}{\sqrt{q} + \sqrt{q - (4/p)}} < \frac{2}{\sqrt{q}} $ and hence it tends zero as $q \to \infty$. Similarly, we have $\beta_{p,q} \geq p\sqrt{q}/2 \geq \sqrt{p}$, where the last inequality follows from $q \geq p \geq 2$.

Combining these observations, we obtain

Lemma 1. $g_{p, q}$ is strictly increasing on $I_{p,q} := [\alpha_{p,q}, \sqrt{p}]$.

Next we investigate the fixed point of $g_{p,q}$. By a brutal-force computation, we find that $g_{p,q}$ has the unique fixed point

$$ \bbox[#fff9e3,border:1px solid #ffeb8e,12px]{ x_{p,q} := \sqrt{\frac{p}{\sqrt{pq} - 1}} } \tag{2}$$

on $(0, \infty)$. Moreover, since $g_{p,q}(\alpha_{p,q}) = \frac{2}{\sqrt{q}} > \alpha_{p,q}$ and $g_{p,q}(\sqrt{p}) = f_q(2/\sqrt{p}) \leq \sqrt{p}$, IVT tells that $x_{p,q} \in I_{p,q}$. So it follows that

Lemma 2. We have

(i) If $x \in [\alpha_{p,q}, x_{p,q})$, then $x < g_{p,q}(x) < x_{p,q}$.

(ii) If $x \in (x_{p,q}, \sqrt{p}]$ then $x > g_{p,q}(x) > x_{p,q}$.

This situation is summarized in the following graph:

$\hspace{7em}$ Graph of the function $g_{p,q}$

We also need a bit technical observation.

Lemma 3. Let $q' \geq p' \geq 2$ be such that $q' \geq q$. Then $g_{p,q}(I_{p,q}) \subseteq I_{p',q'}$.

In order to prove this, it is enough to notice that $ \alpha_{p',q'} \leq \frac{2}{\sqrt{q'}} \leq \frac{2}{\sqrt{q}} = g_{p,q}(\alpha_{p,q}) $ and that $g_{p,q}(\sqrt{p}) = 2/\sqrt{p} \leq \sqrt{2} \leq \sqrt{p'}$.

Finally we need an input from number theory. Let $p_n$ be the $n$-the smallest prime number. Then

Lemma 4. $\lim_{n\to\infty} p_{n+1} / p_n = 1$.

Proof. It is equivalent to saying that the prime gap satisfies $(p_{n+1} - p_n)/p_n \to 0$. See this for the reference.


2. Main proof

Let $x_n = x_{p_n,p_{n+1}}$, where the right-hand side is the symbol defined as $\text{(2)}$. In a similar fasion, we write $g_n = g_{p_n,p_{n+1}}$ and $I_n = I_{p_n,p_{n+1}}$. Then by Lemma 3, we know that $g_n(I_n) \subseteq I_{n+2}$. Also we have $a_2 = \frac{4}{3} \in [2/\sqrt{5}, \sqrt{3}] = I_2$ and $a_3 = \frac{61}{60} \in [2/\sqrt{7}, \sqrt{5}] = I_3$. So by the induction together with the recurrence relation $a_{n+2} = g_n(a_n)$ tells that $a_n \in I_n$ for all $n \geq 2$.

Now by Lemma 4, we know that $x_n \to 1$ as $n\to\infty$. So if we fix a sufficiently small $\epsilon > 0$, there exists $N = N(\epsilon) \geq 2$ such that $|x_n - 1| < \epsilon$ for all $n \geq N$. So by Lemma 1,

  • If $a_n \leq 1-\epsilon$, then $a_n < x_n$ and hence $a_n < a_{n+2} < x_n < 1+\epsilon$.
  • If $a_n \geq 1+\epsilon$, then $a_n > x_n$ and hence $a_n > a_{n+2} > x_n > 1-\epsilon$.
  • Since $a_{n+2}$ always lies between $a_n$ and $x_n$, if $|a_n - 1| < \epsilon$ then $|a_{n+2} - 1| < \epsilon$ as well.

Now let us fix $r \in \{1,2\}$. Then the above observation tells that we have the following trichotomy for the sequence $(a_{2n+r})_{n=0}^{\infty}$.

Case 1. There exists $\epsilon > 0$ such that $a_{2n+r} \leq 1-\epsilon$ for all large $n$. In this case, the sequence is eventually monotone increasing and hence converges.

Case 2. There exists $\epsilon > 0$ such that $a_{2n+r} \geq 1+\epsilon$ for all large $n$. By a similar reasoning, the sequence converges.

Case 3. For all $\epsilon > 0$ we have $|a_{2n+r} - 1| < \epsilon$ for all large $n$. This immediately translates to the statement that $a_{2n+r}$ converges to $1$.

Combining altogether, we have proved that

Proposition. Both $(a_{2n})$ and $(a_{2n+1})$ converge.

Remark. Calibrating the initial value, this proof should work for any sequence $(p_n)$ such that $p_n \geq 2$ and $p_n \nearrow \infty$ with $p_{n+1}/p_n \to 1$. I strongly suspect that we may find some sequence $(p_n)$ such that all these conditions are met but the limit of $(a_{2n})$ and $(a_{2n+1})$ do not coincide. (Geometric sequence might be such a candidate.) So we indeed need more input to settle down the issue of convergence of $(a_n)$.


3. Numerical computation

Here is a computation of first $10^5$ terms using Mathematica:

$\hspace{4em}$ Graph of $a_n$ up to $n = 10^5$

Although it is not easy to read out the convergent bahavior, it seems that the error decays at least as fast as the speed of $1/\log n$. Scaling up the error by log factor indeed provides a better picture:

$\hspace{8em}$ enter image description here

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  • $\begingroup$ very interesting. I'll read it carefully. Sorry for posting just a very partial disussion. I noticed your answer only after pasting mine. $\endgroup$ – dfnu Oct 2 '17 at 13:52
  • $\begingroup$ @Matteo, There is absolutely no need for an apology! As the problem seems hard, we should appreciate any idea and input that we may come up with. Moreover I skimmed over your answer and got an impression that we are working based on almost the same idea. Even some estimations are quite close. It makes me believe that we are on track. :) $\endgroup$ – Sangchul Lee Oct 2 '17 at 13:58
  • $\begingroup$ @Matteo here all contributions are welcome, and I am happy if you and Sangchul are working to help about this problem, it is a nice reference for all users. Many thanks to you and Sangchul. $\endgroup$ – user243301 Oct 2 '17 at 18:55
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Download Pari/GP and run this code

 N = 10^5;
 p = primes(N);
 a = vector(N); a[1]=1.0; /* 1.0 means float, with a[1] = 1; instead the sequence of rationals is computed in closed form */
 for( k = 1,  N-1, { a[k+1] = a[k]/p[k+1]+1/a[k];});

 x = vector(N); for( k = 1,  N, { x[k] = k;});
 plothraw(x,a);
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The answer given in the meantime is very iteresting. However, as you suggested, I write my own results as answer instead of comment, even though I am far from a conclusion. I would start from an "easier" situation, with the sequence $$ a_{k+1} = \frac{a_k}{M} + \frac{1}{a_k},$$ with $M>2$. I now analyse the function $$y_M(x) = \frac{x}{M}+\frac{1}{x}$$ and interesect it with $y=x$ to look for the equilibrium point which occurs at $$x_C= \sqrt\frac{M}{M-1}.$$ The function reaches its minimum at $$x_m = \sqrt M.$$ If you take $x=f_M(x_c)=\frac{2\sqrt M}{M}$ and iterate the sequence once more, you get the point $$ f_M\left(\frac{2\sqrt M}{M}\right)=\sqrt M\left(\frac{2}{M^2}+\frac{1}{2}\right) =x_l $$ that, for $M>2$ is less then $\sqrt M$. As a consequence, once a value $$ \frac{2\sqrt{M}}{M} \leq a_k \leq \sqrt M\left(\frac{2}{M^2}+\frac{1}{2}\right)\tag{1}\label{eq:one} $$ is reached, the sequence values remain "trapped" in the range $$ \frac{2\sqrt{M}}{M} <a_{k+1}< \sqrt M$$ and converge to $x_C$. So:

  1. When the starting point $a_1$ is in the range $$f^{-1}_M(x_l)<a_k<x_l,$$ where the left extremity is taken to be the smallest between the two solutions, then the sequence will oscillate with decreasing amplitude and converge to $x_C$.
  2. If $a_1 > x_l$ then the sequence will first decrease monotonically until the range \eqref{eq:one} is reached; after that the it will behave as in 1.
  3. Finally, if $a_1 < f^{-1}_M(x_l)$, then from $a_2$ onwards, the sequence will behave as in 2.

As I said, no conclusions here regarding your sequence, in which $M$ is a (growing) function of $k$. The consequences are, I believe:

a. Affecting the ranges at each iteration. b. As $M$ increases, the sequences approaches the behavior of the sequence $a_k=\frac{1}{a_k}$, which of course does not converge.

That is all for now.

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  • $\begingroup$ Many thanks I am going to study yours and Sangchul's answer. $\endgroup$ – user243301 Oct 2 '17 at 18:58
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Assume that

$1-\frac{1}{p_k} \geq a_k \geq 1-\frac{1}{k} \geq 0.8$,so $k \geq 5$.

We will prove that $1-\frac{1}{p_{k+2}} \geq a_{k+2} \geq 1-\frac{1}{k+2}$.

We need first to bound $a_{k+1}$ by upper and lower bound.

$a_{k+1} = \frac{a_k}{p_{k+1}} +\frac{1}{a_k}$, so $\frac{d}{d a_k} (\frac{a_k}{p_{k+1}} +\frac{1}{a_k}) = \frac{1}{p_{k+1}}-\frac{1}{a_k^2} < 0$ since $1 \geq a_k \geq 0.8$ and $p_{k+1} \geq 13$, which means that an upper bound for $a_{k+1}$ is when $a_k = 1-\frac{1}{k}$ and lower bound is when $a_k = 1-\frac{1}{p_k}$.

Substituting the values for $a_k = 1-\frac{1}{k} , 1-\frac{1}{p_k}$ in the equation $a_{k+1} = \frac{a_k}{p_{k+1}}+\frac{1}{a_k}$ gives the following upper and lower bound for $a_{k+1}$.

$1+\frac{1}{p_k-1} +\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}} \leq a_{k+1} \leq 1+\frac{1}{k-1}+\frac{1}{p_{k+1}}-\frac{1}{k p_{k+1}}$.

The same idea to bound $a_{k+2} = \frac{a_{k+1}}{p_{k+2}} +\frac{1}{a_{k+1}}$ gives us the following upper and lower bound.

$ \frac{1+\frac{1}{k-1}+\frac{1}{p_{k+1}}-\frac{1}{k p_{k+1}}}{p_{k+2}}+\frac{1}{1+\frac{1}{k-1}+\frac{1}{p_{k+1}}-\frac{1}{k p_{k+1}}} \leq a_{k+2} \leq \frac{1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}}{p_{k+2}}+ \frac{1}{1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}}$

Now we just need to prove that,

$ \frac{1+\frac{1}{k-1}+\frac{1}{p_{k+1}}-\frac{1}{k p_{k+1}}}{p_{k+2}}+\frac{1}{1+\frac{1}{k-1}+\frac{1}{p_{k+1}}-\frac{1}{k p_{k+1}}} \geq 1-\frac{1}{k+2} $, and

$\frac{1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}}{p_{k+2}}+ \frac{1}{1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}} \leq 1-\frac{1}{p_{k+2}} $

First,

$\frac{1+\frac{1}{k-1}+\frac{1}{p_{k+1}}-\frac{1}{k p_{k+1}}}{p_{k+2}}+\frac{1}{1+\frac{1}{k-1}+\frac{1}{p_{k+1}}-\frac{1}{k p_{k+1}}} \geq \frac{1+\frac{1}{k-1}}{p_{k+2}} +\frac{1}{1+\frac{1}{k-1}+\frac{1}{p_{k+1}}} \geq 1-\frac{1}{k+2}$

So $ \frac{1}{p_{k+2}}+\frac{1}{(k-1)p_{k+2}} +\frac{1}{\frac{k}{k-1}+\frac{1}{p_{k+1}}} \geq \frac{k+1}{k+2}$, multiply by $p_{k+2}$ to get

$1+\frac{1}{k-1} +\frac{p_{k+2}}{\frac{k p_{k+1}+1}{(k-1)p_{k+1}}} \geq \frac{k+1}{k+2} p_{k+2}$, multiply by $(k-1)$, to get

$ k-1+1 +\frac{p_{k+2}(k-1)(k-1)p_{k+1} }{k p_{k+1}+1} \geq \frac{(k+1)(k-1)}{k+2} p_{k+2} $ which is $ k +\frac{p_{k+2}(k-1)(k-1)p_{k+1} }{k p_{k+1}+1} \geq \frac{(k+1)(k-1)}{k+2} p_{k+2} $

multiply by $ (k+2)(k p_{k+1}+1)$, to get

$k(k+2)(k p_{k+1}+1) + (k+2) p_{k+2} p_{k+1} (k-1)^2 \geq p_{k+2} (k^2-1) (k p_{k+1} +1) $, which is

$k(k+2)(k p_{k+1} +1) \geq p_{k+2}((k^2-1) (k p_{k+1}+1)-p_{k+1}(k+2)(k-1)^2)$, which is

$(k^2+2k)(k p_{k+1}+1) \geq p_{k+2}(2 k p_{k+1}-2p_{k+1}+k^2-1)$,

We have that $k^2+2k \geq k^2$, so we arrive at

$ k^2 (k p_{k+1} +1) \geq k^2 (k p_{k+1}) \geq p_{k+2}(2 k p_{k+1}+k^2) \geq p_{k+2}(2 k p_{k+1}-2p_{k+1}+k^2-1)$, which is

$ k^3 p_{k+1} \geq p_{k+2}(2 k p_{k+1} +k^2)$

We know that $k^2 \leq k p_{k+1}$, so $k^3 p_{k+1} \geq p_{k+2}(2 k p_{k+1}+k p_{k+1}) \geq p_{k+2}(2 k p_{k+1} +k^2)$, which is

$ k^3 p_{k+1} \geq 3 k p_{k+2} p_{k+1}$ , which is $k^3 \geq 3 p_{k+2}$, which is

$ k^2 \geq 3p_{k+2}$ which is true for all $k \geq 12$.

Second part,

$\frac{1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}}{p_{k+2}}+ \frac{1}{1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}} \leq 1-\frac{1}{p_{k+2}} $, multiply by $p_{k+2}$, we get that

$1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}+ \frac{p_{k+2}}{1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}} \leq p_{k+2}-1 $,

multiply by $ 1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}$ to get that

$ (1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}})^2+p_{k+2} \leq p_{k+2}(1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}})-(1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}})$

At least $p_{k+2} \geq p_k +6$ (Hint : no $3$ consecutive twin primes) and $p_{k+2} \geq p_{k+1}+2$, so

$(1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}})^2+(1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}) \leq p_{k+2}(\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}) $ , by the idea above we get that

$(1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}})^2+(1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}) \leq (\frac{p_k+6}{p_{k-1}}+\frac{p_{k+1}+2}{p_{k+1}}-\frac{p_{k+2}}{p_k p_{k+1}}) $, we get that

$(1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}})^2+(1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}}-\frac{1}{p_k p_{k+1}}) \leq (1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}})^2+1+\frac{1}{p_{k-1}}+\frac{1}{p_{k+1}} \leq (\frac{p_k+6}{p_{k-1}}+\frac{p_{k+1}+2}{p_{k+1}}-\frac{p_{k+2}}{p_k p_{k+1}}) $

Its easy to prove that $\frac{1}{p_{k}-1} +\frac{1}{p_{k+1}} \leq \frac{1}{p_k-1}+\frac{1}{p_k+2} \leq \frac{2}{p_k}$ for all $k \geq 3$.

$(1+\frac{2}{p_k})^2+1+\frac{2}{p_k} \leq 1+\frac{7}{p_k-1}+1+\frac{2}{p_{k+1}}-\frac{p_{k+2}}{p_k p_{k+1}}$, which is

$1+\frac{4}{p_k}+\frac{4}{p_k^2}+1+\frac{2}{p_k} \leq 1+\frac{7}{p_k}+1+\frac{2}{p_{k+1}} -\frac{p_{k+2}}{p_k p_{k+1}} \leq 1+\frac{7}{p_k-1}+1+\frac{2}{p_{k+1}} -\frac{p_{k+2}}{p_k p_{k+1}} $

Which is just $\frac{4}{p_k^2}+\frac{p_{k+2}}{p_k p_{k+1}} \leq \frac{1}{p_k}+\frac{2}{p_{k+1}}$

There is a prime between $[x,\frac{6}{5}x]$ for all $x \geq 25$,so

$\frac{4}{p_k^2} +\frac{1.44p_k}{p_k^2} \leq \frac{1}{p_k}+\frac{2}{1.2p_k}$, which is $ \frac{4}{p_k^2} +\frac{1.44}{p_k} \leq \frac{8}{3p_k}$ which is true for all $k \geq 10$.

Thus concluding the proof for all $k\geq 12$ but we need to check for small $k$, checking for small $k$ we get that its true for all $k\geq 19$ and $k$ is odd.

And since $a_k$ for even $k$ converges according to $a_k$ for $k$ odd, then we can conclude that $\lim \limits_{k \to \infty} a_k =1$.

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  • $\begingroup$ it bugged me that such a beautiful question with no complete answer, so i was determined that first thing in the morning is sitting down and writing a proof for the question (actually not the first thing, first i listened to Firooz morning songs, but then right away i start solving the problem, i really enjoyed solving it. $\endgroup$ – Ahmad Oct 16 '17 at 7:55
  • $\begingroup$ Many thanks, my skills in mathematics aren't the best thus I am going to need some days to read your proof. Many thanks for your attention and work. $\endgroup$ – user243301 Oct 16 '17 at 9:02
  • $\begingroup$ you are very welcome, you won't believe how much effort i put to the answer, i just could not agree that such amazing question with no full answer, i kept bugging me, until today :) $\endgroup$ – Ahmad Oct 16 '17 at 18:51

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