Test for Cauchy sequence (with limsup and log)

In summary, a Cauchy sequence is a sequence of numbers that progressively get closer together. Testing for Cauchy sequences is important in determining convergence and the limit of the sequence. The limsup of a Cauchy sequence is the largest value approached by the terms in the sequence. To test for Cauchy sequences using the limsup, the limit superior must equal 0. Logarithms can be used to make it easier to determine if a sequence is a Cauchy sequence or not by compressing the sequence.
  • #1
Kokuhaku
9
0
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.
 
Physics news on Phys.org
  • #2
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
$$\limsup_{n\to \infty} \frac{1}{n}\log\, \lvert x_{n+1} - x_n\rvert \ge \limsup_{k\to \infty} \frac{1}{n_k}\log\, \lvert x_{n_{k+1}} - x_{n_k}\rvert \ge \limsup_{k\to \infty} \frac{1}{n_k}\log M = 0.$$
 
  • #3
Kokuhaku said:
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.
 
  • #4
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?
 
  • #5
Kokuhaku said:
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 $\varepsilon' < \varepsilon$.
 
  • #6
Kokuhaku said:
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 $a + \varepsilon < 0$.
 

FAQ: Test for Cauchy sequence (with limsup and log)

1. What is a Cauchy sequence?

A Cauchy sequence is a sequence of numbers that gets closer and closer together as the sequence progresses. This means that for any given value, there exists a point in the sequence after which all the subsequent terms are within a certain distance from that value.

2. Why is it important to test for Cauchy sequences?

Testing for Cauchy sequences can help us determine if a given sequence of numbers converges or not. This is important because it allows us to determine the limit of the sequence, which can be useful in many mathematical and scientific applications.

3. What is the limsup of a Cauchy sequence?

The limsup (limit superior) of a Cauchy sequence is the largest value that the terms in the sequence approach as the sequence progresses. It is used to determine the convergence of a sequence and can be seen as a sort of "upper bound" for the sequence.

4. How do you use the limsup to test for Cauchy sequences?

To test for a Cauchy sequence using the limsup, we first find the limit superior of the sequence. If the limit superior is equal to 0, then the sequence is a Cauchy sequence. However, if the limit superior is greater than 0, then the sequence is not a Cauchy sequence and does not converge.

5. What is the role of logarithms in testing for Cauchy sequences?

In some cases, taking the logarithm of a sequence can make it easier to determine if it is a Cauchy sequence or not. This is because the logarithm can "compress" the sequence and make it easier to see if the terms are getting closer together or not.

Back
Top