Test for Cauchy sequence (with limsup and log)

[Kokuhaku]

If $\{x_n\}_{n \ge 1}$ is real sequence and $\limsup\limits_{n \to \infty} \frac{1}{n} \log |x_{n+1}-x_n|<0$, prove that $\{x_n\}$ is Cauchy sequence.

My work: Let $a=\limsup\limits_{n \to \infty} \frac{1}{n} \log |x_{n+1}-x_n| <0$. Then, for every $\varepsilon >0$ there exist $N \in \mathbb{N}$ such that for all $n \ge N$ is $\frac{1}{n} \log |x_{n+1}-x_n| < a+ \varepsilon < \varepsilon$. From that, we see that for $n \ge N$ we have $|x_{n+1}-x_n| < e^{n \varepsilon}$.

Then I tried to use $$|x_m - x_n| \le \sum_{k=n}^{m-1} |x_{k+1} - x_k| \le \sum_{k=n}^{m-1} e^{k \varepsilon} = \frac{e^{\varepsilon m} - e^{\varepsilon n}}{e-1}$$ for $m,n \ge N$.

Now, problem is that $e^{\varepsilon m} - e^{\varepsilon n}$ can be arbitrary big, so probably this isn't the best way to solve this problem.

 

[Euge]

Hi Kokuhaku,

A proof by contraposition seems appropriate here. Suppose $\{x_n\}_{n=1}^\infty$ is not a Cauchy sequence. Then there exists a positive number $M$ and an increasing sequence $n_1 < n_2 < n_3 < \cdots$ of indices such that $\lvert x_{n_{k+1}} - x_{n_k}\rvert \ge M$, for $k = 1,2,3,\ldots$. Consequently
$$\limsup_{n\to \infty} \frac{1}{n}\log\, \lvert x_{n+1} - x_n\rvert \ge \limsup_{k\to \infty} \frac{1}{n_k}\log\, \lvert x_{n_{k+1}} - x_{n_k}\rvert \ge \limsup_{k\to \infty} \frac{1}{n_k}\log M = 0.$$

 

[Opalg]

If $\{x_n\}_{n \ge 1}$ is real sequence and $\limsup\limits_{n \to \infty} \frac{1}{n} \log |x_{n+1}-x_n|<0$, prove that $\{x_n\}$ is Cauchy sequence.

My work: Let $a=\limsup\limits_{n \to \infty} \frac{1}{n} \log |x_{n+1}-x_n| <0$. Then, for every $\varepsilon >0$ there exist $N \in \mathbb{N}$ such that for all $n \ge N$ is $\frac{1}{n} \log |x_{n+1}-x_n| < a+ \varepsilon < \varepsilon$. From that, we see that for $n \ge N$ we have $|x_{n+1}-x_n| < e^{n \varepsilon}$.

Then I tried to use $$|x_m - x_n| \le \sum_{k=n}^{m-1} |x_{k+1} - x_k| \le \sum_{k=n}^{m-1} e^{k \varepsilon} = \frac{e^{\varepsilon m} - e^{\varepsilon n}}{e-1}$$ for $m,n \ge N$.

Now, problem is that $e^{\varepsilon m} - e^{\varepsilon n}$ can be arbitrary big, so probably this isn't the best way to solve this problem.

It may seem too obvious to be worth mentioning, but the fact that $a=\limsup\limits_{n \to \infty} \frac{1}{n} \log \bigl|x_{n+1}-x_n\bigr| <0$ tells you that $a$ is negative. So you can choose $\varepsilon$ sufficiently small that $a + \varepsilon$ is still negative.

I prefer to write $c$ for $a + \varepsilon.$ Then $c<0$. From the fact that $\frac{1}{n} \log \bigl|x_{n+1}-x_n\bigr| < a+ \varepsilon = c$ it follows that $\bigl|x_{n+1}-x_n\bigr| < e^{nc}$ (for all sufficiently large $n$). Then (just as in your post) $$\bigl|x_m - x_n\bigr| \leqslant \sum_{k=n}^{m-1} \bigl|x_{k+1} - x_k\bigr| < \sum_{k=n}^{m-1} e^{kc} = \frac{e^{mc} - e^{nc}}{e^c-1}$$ for $m > n \geqslant N$. But now the fact that $c$ is negative tells you that the numerator can be made as small as you want provided that $n$ and $m$ are sufficiently large.

 

[Kokuhaku]

I see that Opalg, but that works only for $a+\varepsilon <0$, that is $\varepsilon < -a$. But $\varepsilon$ can be arbitrary. Am I missing something?

 

[Euge]

That works only for $a+\varepsilon <0$, that is $\varepsilon < -a$. But $\varepsilon$ can be arbitrary. Am I missing something?

Since the argument works for $\varepsilon < -a$, it also works for any greater value of $\varepsilon$: Given $\varepsilon > 0$, let $\varepsilon' > 0$ such that $\varepsilon' < -a$ and $\varepsilon' < \varepsilon$.

 

[Opalg]

I see that Opalg, but that works only for $a+\varepsilon <0$, that is $\varepsilon < -a$. But $\varepsilon$ can be arbitrary. Am I missing something?

If you are given a problem where you need to prove that a lim sup exists, then you have to show that the defining property is satisfied for all $\varepsilon>0$. But in this problem, you are told that the lim sup exists. So you know that the defining property holds for all $\varepsilon>0$. In order to solve the problem, it is sufficient in this case to choose one convenient value of $\varepsilon$, namely one for which $a + \varepsilon < 0$.