# $a_{n+1}=a_n^{n+1}+a_n+1$

Let $a_1>1$ and $$a_{n+1}=a_n^{n+1}+a_n+1, \: \forall n \geq 1$$ Edit: Prove that there is a real number $x \neq 0$ such that $$\lim_{n \to \infty}\frac{a_n}{x^{n!}}=1$$

I think that $x=a_1$ works, and I tried to use the Squeeze Theorem to prove that. From $a_{n+1}>a_n^{n+1}>a_{n-1}^{n(n+1)}>\dots>a_1^{(n+1)!}$ we get that $1<\frac{a_{n+1}}{a_1^{(n+1)!}}$ and then I tried to find another bound for $a_{n+1}$ such that $\frac{a_{n+1}}{a_1^{(n+1)!}}<b_{n+1} \to 1$, but nothing seems to work.

• No, that doesn't work, and the formulation is misleading. You can show the existence of such an $x$, but you can't find it without computing the whole sequence. – Professor Vector Jan 12 '18 at 11:54
• Okay, I'll edit the statement! – Shroud Jan 12 '18 at 11:55

## 2 Answers

Let

$$x = \lim_{n\to\infty} a_n^{1/n!}.$$

We know from @frame95's answer that $x$ exists. Now set $\delta_n = \log a_n - n! \log x$ and notice that

$$\frac{\delta_{n+1}}{(n+1)!} = \frac{\delta_n}{n!} + \frac{1}{(n+1)!}\log\left( 1 + \frac{1}{a_n^n} + \frac{1}{a_n^{n+1}} \right).$$

Since we know that $\delta_n/n! \to 0$, recursively applying the above relation and taking limit shows that

$$\frac{\delta_n}{n!} = -\sum_{k = 0}^{\infty} \frac{1}{(n+k+1)!}\log\left( 1 + \frac{1}{a_{n+k}^{n+k}} + \frac{1}{a_{n+k}^{n+k+1}} \right)$$

Now by utilizing simple estimations $a_{n+k} \geq a_n^{k!} \geq a_1^{k!}$ and $\log(1+t) \leq t$ for $t \geq 0$, we obtain the following very crude bound

$$\left| \delta_n \right| \leq \sum_{k=0}^{\infty} \frac{n!}{(n+k+1)!} \cdot 2 a_1^{-k!}.$$

Since each summand is bounded by $2a_1^{-k!}$ and $\sum_{k=0}^{\infty} 2a_1^{-k!} < \infty$, dominated convergence theorem tells that

$$\limsup_{n\to\infty} \left| \delta_n \right| \leq \sum_{k=0}^{\infty} \lim_{n\to\infty} \frac{n!}{(n+k+1)!} \cdot 2 a_1^{-k!} = 0.$$

Therefore $\delta_n \to 0$ and hence $a_n!/x^{n!} = e^{\delta_n} \to 1$ as required.

• Wow! Good (team) job guy :) very neat proof! – Andrea Marino Jan 13 '18 at 10:36

The guess is wrong, but if you go a step further... Let's stop the chain of inequalities a bit before: $$a_n> a_{n-1}^n > \ldots > a_2 ^{n!/2} = (\sqrt{a_2} )^{n!}$$ Is this a better estimate? Yes, because $\sqrt{a_2 }=\sqrt{ a_1^2 + a_1+1} > a_1$. Let's call $d_r = a_r^{1/r!}$. Generalizing this argument, we get $$a_{r+k} > a_{r+k-1}^{r+k} > \ldots > a_r^{ (r+k)!/r!} = d_r^{(r+k)!}$$ Let's analyze $\{d_r\}$. We will show that it is an increasing and bounded sequence, thus convergent to a number $\ell$. Infact $$(1) \ \ increasing \ \ \ d_{r+1} = a_{r+1}^{1/(r+1)!} = ( a_r^{r+1} +a_r+1)^{1/(r+1)!} > d_r$$ By induction, one can easily show that $a_n +1 < (a_1+1) a_n^{n+1}$. Thus, $a_{n+1} < (a_1+2)a_n^{n+1}$. This yields $$(2) \ \ bounded \ \ \ d_{n+1} < (a_1+2)^{1/(n+1)!} d_n < 2^{\sum^{n+1} (1/k!)} a_1/2 < 2^{e-1} a_1$$ If $x$ exists, then $x=\ell$. Infact $$\lim_{n \to \infty} \frac{a_n}{x^{n!}} = \lim_{n \to \infty} \left ( \frac{d_n}{x} \right )^{n!} = 1 \ \Rightarrow \ \frac{\ell}{x} = \lim_{n \to \infty} \frac{d_n}{x} = 1$$ Here my results stop: I can't show the refined result that this $x$ exist!