A surprising inequality about a $\limsup$ for any sequence of positive numbers

Question

Prove the following:

Let $\{x_n\}_{n=1}^{\infty}$ be a sequence of positive real numbers. Then $$ \limsup \frac{x_1+x_2+...+x_n+x_{n+1}}{x_n} \geq 4 \ .$$

Note:

1. Yes, $4$ is sharp: $x_n=2^n$ sequence.
2. I have already proved that $$ \liminf x_n < \infty \implies \limsup \frac{x_1+x_2+...+x_n+x_{n+1}}{x_n} = \infty.$$ So, the interesting case is when $\lim x_n = \infty$.

Answer 1

Solution(as in my lecture)
Suppose that for a positive constant $c$ and a number $N$ $$ \frac{a_1+\dots a_{n+1}}{a_n} \le c \quad \text{ for all }n \ge N $$ Let $b_n:= \sum_{i=1}^n a_n$, the inequality above is equivalent to $$b_{n+1} \le c(b_{n}-b_{n-1}) \text{ for all }n \ge N$$ This form of the inequality is more or less more easier to deal with, but let's try it further, by letting $c_{n}$ denote $\frac{b_n}{b_{n-1}}$,then $$c_{n+1}+\frac{c}{c_n} \le c$$ Thus $(c_n , n \ge N)$ is a sequence of bounded real numbers, thus we can choose a subsequence $(c_{n_k},k\ge 1)$ such that $$\lim_{k \rightarrow \infty} c_{n_k}= \underbrace{\limsup_{n \rightarrow \infty } c_n}_{=:a}$$ Thus $$c \ge \limsup_{k \rightarrow \infty} \left(c_{n_k}+\frac{c}{c_{n_k-1}} \right)=a+\limsup_{k \rightarrow \infty} \frac{c}{c_{n_k-1}} \ge a+ \frac{c}{ \limsup c_n} =a+\frac{c}{a} \ge 2\sqrt{c} $$ Or, $c$ must be bigger than $4$. This means$$\text{ if} c \ge \limsup \frac{ a_1+\dots+a_{n+1}}{a_n} \text{ then } c \ge 4$$ Hence the conclusion. $\square$

A small story: This question brings back lots of memories. The first time I encounter it was when solving the famous analysis problem books by Kaczor and Nowak. A very long time ago. Then after that, in a small tutor class with gifted highschool students, I told them that it was important to find a good presentation of your problem.
And this question was the example I used in that class.

Answer 2

First, assume that $\lim\sup \frac{x_{1}+\cdots+x_{n+1}}{x_{n}}=L>0$ is finite. Define a sequence of polynomial $\{p_{n}(x)\}$ by $p_{0}(x)=1, p_{1}(x)=x-1$ and \begin{align} p_{n+1}(x)=(x-1)p_{n}(x)-p_{n-1}(x)-\cdots-p_{1}(x)-p_{0}(x).\,\,(n\geq 1) \end{align}

Claim. $p_{n}(L)>0$ for all $n\geq 1$.

proof. We have $x_{1}+x_{2}+\cdots+x_{n+1}\leq Lx_{n}$ for all $n\geq 1$. (Edited : Actually, we only have this for sufficiently large $n$. However, this can be modified to all $n$ with a user254665's comment.) By induction, we will prove the following : \begin{align} x_{n+1}\leq \frac{p_{n}(L)}{p_{n-1}(L)}x_{n} \end{align} (If we prove this inequality, then claim immediately follows from $x_{n+1}, x_{n}>0$.) Since $x_{1}+x_{2}\leq Lx_{1}$, it holds for $n=1$. If it holds for $n\leq N-1$, then \begin{align} x_{N+1}&\leq (L-1)x_{N}-x_{N-1}-x_{N-2}-\cdots-x_{1} \\ &\leq \left(L-1-\frac{p_{N-2}(L)}{p_{N-1}(L)}-\frac{p_{N-2}(L)}{p_{N-1}(L)}\frac{p_{N-3}(L)}{p_{N-2}(L)}-\cdots-\frac{p_{N-2}(L)}{p_{N-1}(L)}\frac{p_{N-3}(L)}{p_{N-2}(L)}\cdots \frac{p_{0}(L)}{p_{1}(L)}\right)x_{N} \\ &=\frac{p_{N}(L)}{p_{N-1}(L)}x_{N}. \end{align}

Now using the recurrence relation, we can prove $p_{n+1}(x)=x(p_{n}(x)-p_{n-1}(x))$ by subtracting two recurrence relations for $n$ and $n-1$. Then by using characteristic polynomial method, we get \begin{align} p_{n}(x)=\frac{\sqrt{x^{2}-4x}+x-2}{2\sqrt{x^{2}-4x}}\left(\frac{x+\sqrt{x^{2}-4x}}{2}\right)^{n}+\frac{\sqrt{x^{2}-4x}-x+2}{2\sqrt{x^{2}-4x}}\left(\frac{x-\sqrt{x^{2}-4x}}{2}\right)^{n} \end{align} for $x\neq 4$.

Now, suppose $0<L<4$. Let \begin{align} \frac{L+\sqrt{L^{2}-4L}}{2}=\frac{L+i\sqrt{4L-L^{2}}}{2}=\sqrt{L}e^{i\theta} \end{align} where $\cos \theta=\frac{\sqrt{L}}{2}<1$. By direct computation, we have \begin{align} p_{n}(L)=L^{n/2}\left(\cos n\theta+\frac{L-2}{\sqrt{4L-L^{2}}}\sin n\theta\right)=\frac{2L^{n/2}}{\sqrt{4L-L^{2}}} \sin (n\theta+\alpha) \end{align} where $\cos\alpha=\frac{L-2}{2}$. Since $0<\theta<\pi/2$, we can choose $n$ s.t. $\sin (n\theta+\alpha)<0\Rightarrow p_{n}(L)<0$. Contradiction. So we have $L\geq 4$.

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There is an easier proof which is discovered by my friend. Let $S_{n}=x_{1}+\cdots + x_{n}$ and $t_{n}=S_{n}/S_{n+1}$. Then $0<t_{n}<1$. Suppose $\limsup\frac{x_{1}+\cdots+x_{n+1}}{x_{n}}=L<4$. Then we have \begin{align} \frac{x_{1}+\cdots+x_{n+1}}{x_{n}}\leq L\Leftrightarrow t_{n}\geq \frac{1}{L(1-t_{n-1})} \end{align} for $n\geq N$ for some $N\in \mathbb{N}$. If we put $f(x)=\frac{1}{L(1-x)}$, $t_{n}\geq f(t_{n-1})$. We can check that $f(x)>x$ and $t_{n}$ diverges, i.e. $t_{n}>1$ for some $n$. Contradiction.