6

Define a sequence $\{a_n\}_{n=1}^{\infty}$ satisfying $\displaystyle a_1>1$ and $\frac{a_{n+1}}{a_n}=1+\frac{1}{\ln a_n}.$ Find $\displaystyle\lim\limits_{n \to \infty}\frac{a_n}{e^{\sqrt{2n}}}$.

First, it is easy to observe that the sequence $\{a_n\}$ is monotonically increasing. Therefore, it either converges to some finite limit $1<L<+\infty$, or diverges to $+\infty$. However, the former scenario is impossible. If we assume convergence, taking the limit of the recurrence relation would yield $L=L+\frac{L}{\ln L}$ ​, which has no solution. Thus, the only possibility is that the sequence diverges to $+∞$, i.e., $a_n \to +\infty$.

Now, applying Stolz theorem, we obtain $$ \lim_{n \to \infty}\frac{a_n}{{\rm e}^{\sqrt{2n}}}=\lim_{n \to \infty}\frac{a_{n+1}-a_n}{{\rm e}^{\sqrt{2(n+1)}}-{\rm e}^{\sqrt{2n}}}=\lim_{n \to \infty}\frac{\frac{a_n}{\ln a_n}}{{\rm e}^{\sqrt{2n}}({\rm e}^{\sqrt{2n+2}-{\sqrt{2n}}}-1)}.$$

This would help? How to go on ?

Edit: One kind of solution is posted here, but I can't understand it well.

metamorphy
  • 43,591
WuKong
  • 14,376
  • 1
  • 18
  • 45

2 Answers2

4

Let $u_n=\frac{1}{\log(a_n)}$, we have $\begin{cases}u_{n+1}=f(u_n)\\u_0=x\end{cases}$ where $f(x)=(x^{-1}+\log(1+x))^{-1}$

This function has expansion $f(x)=x+ax^{p}+O(x^{p+1})$ and the sequence converges $u_n\to0$ , thus we adapt a general technique from this answer on MO (recommend reading for explanation, I will just summarize the main result), that is we need to find the expansion of Fatou coordinate $\alpha(x)$ where $\alpha(f(x))=\alpha(x)+1$ , this can be done via expansion of $1/\alpha'(x)$. In general, we have the result: $$\alpha(x)=\lim_{n\to\infty}\left(\alpha(u_n)-n\right)$$ By symbolic computation: $$f(x)=(x^{-1}+\log(1+x))^{-1}\implies \alpha(x)\sim\frac{1}{2x^2}+\frac{1}{2x}+\frac{7}{12}\log x+O(1)$$ Let $b_n=1/u_n$, we have: $$n\sim G(b_n)=\frac{1}{2}b_n^2+\frac{1}{2}b_n-\frac{7}{12}\log b_n+O(1)$$ The idea is that we just substitute $n$ into the limit. The limit in question: $$\log L=\lim_{n\to\infty}\left(b_n-\sqrt{2n}\right)$$$$=\lim_{n\to\infty}\left(b_n-\sqrt{2\left(\frac{1}{2}b_n^2+\frac{1}{2}b_n-\frac{7}{12}\log b_n+O(1)\right)}\right) $$$$=\lim_{x\to\infty}\left(x-\sqrt{x^2+x-\frac{7}{6}\log x+O(1)}\right)=\frac{-1}{2}$$ The answer is $$L=\lim_{n\to\infty}\frac{a_n}{e^{\sqrt{2n}}}=e^{-\frac{1}{2}}$$ Below is the Mathematica code I wrote to calculate $G(x)$ (or $\alpha(x)$) and also the limit in question:

(* sequence b(n+1)=myFun(b(n)), b(n)->0, myFun(x)=x+O(1) "Laurent series"*)
(* AbelAsymptotic return asymptotic G where n∼G(b(n))+const *)

AbelAsymptotic[H_,n_]:=Module[{F,f,g,const},
F[x_]=Asymptotic[H[y]-y,y->0]/.y->x;
pow = Limit[Log[Abs[F[x]]]/Log[x],x->Infinity];
f[0,x_]=F[x];
Do[f[x_,c_]=f[i-1,x]+cx^(pow+i);
g=Solve[(Asymptotic[f[H[x],c]-f[x,c]D[H[x],x],x->0]/.x->1)==0,c];
const = c/.g[[1]];
f[i,x_]=f[i-1,x]+constx^(pow+i),{i,pow2+n}];
Simplify[Integrate[Normal[Series[1/f[pow*2+n,x],{x,0,n}]],x]/.(x->1/z),z>0] //Expand
];


myFun[x_]:= x+Log[1+1/x]; (* Input function here *)


fun = Function[x,(myFun[1/x])^(-1)];
AbelAsymptotic[fun,3] (expansion accuracy is after parameter "fun")
(End function, below is code to solve limit problem)


funToCal[x_,y_] = x-Sqrt[2*y]; (the limit x=b(n),y=n)
Limit[funToCal[z,AbelAsymptotic[fun,3]],z->Infinity]
Quý Nhân
  • 2,706
4

Someone posts a solution on a Chinese website 知乎. Now, I translate it into English as follows.

Let $a_n=e^{b_n}$, where the sequence $\{b_n\}$ satisfies $$b_1>0,~~~b_{n+1}=b_n+\ln\left(1+\frac{1}{b_n}\right).$$ First, it is straightforward to prove that $\{b_n\}$ is strictly increasing and diverges to $+\infty$. Applying Stolz's theorem, we obtain \begin{align*} \lim_{n \to \infty}\frac{b_n^2}{2n}=\lim_{n \to \infty}\frac{b^2_{n+1}-b^2_n}{2(n+1)-2n}=\frac{1}{2}\lim_{n \to \infty}\left[2b_n\ln\left(1+\frac{1}{b_n}\right)+\ln^2\left(1+\frac{1}{b_n}\right)\right]=1,\\ \end{align*} thus establishing $b_n\sim \sqrt{2n}$.

Next, observe the recurrence relation $$b_{k+1}=b_k+\ln\left(1+\frac{1}{b_k}\right)=b_k+\frac{1}{b_k}-\frac{1}{2b_k^2}+\cdots.$$ Squaring both sides yields $$b_{k+1}^2-b_k^2=2-\frac{1}{b_k}+\cdots.$$ Summing from $k=1$ to $n$ gives $$b_{n+1}^2-b_1^2=2n-\sqrt{2n}+o(\sqrt{n}).$$ Through asymptotic expansion \begin{align*} b_{n+1} &= \sqrt{2n - \sqrt{2n} + o(\sqrt{n})} \\ &= \sqrt{2n}\sqrt{1 - \frac{1}{\sqrt{2n}} + o\left(\frac{1}{\sqrt{n}}\right)} \\ &= \sqrt{2n}\left(1 - \frac{1}{2}\cdot\frac{1}{\sqrt{2n}} + o\left(\frac{1}{\sqrt{n}}\right)\right) \\ &= \sqrt{2n} - \frac{1}{2} + o(1), \end{align*} it follows that $$ \lim_{n\rightarrow\infty}\left(b_{n} - \sqrt{2n}\right) = \lim_{n\rightarrow\infty}\left(b_{n+1} - \sqrt{2n}\right) - \lim_{n\rightarrow\infty}\left(b_{n+1} - b_{n}\right) = -\frac{1}{2} - 0 = -\frac{1}{2},$$ which implies the required limit is $e^{-\frac{1}{2}}$.

WuKong
  • 14,376
  • 1
  • 18
  • 45