show that $x_{n+1}$- $x_n$ $\rightarrow $ 0 and n $\rightarrow \infty$ does not imply cauchy

Question

Let $(x_n)_n$ be a sequence of real numbers and consider the following statements

a) $(x_n)_n$ is cauchy

b) $x_{n+1}$- $x_n$ $\rightarrow $ 0 and n $\rightarrow \infty$.

show that a) implies b)

by considering $(x_n)_n =(1+\frac{1}{2}+\frac{1}{3}...+\frac{1}{n})_n$ show that b) does not imply a).

Answer 1

$x_{n+1}-x_{n}=1/n\rightarrow 0$ but for $n>m$, $x_{n}-x_{m}=\dfrac{1}{m+1}+\cdots+\dfrac{1}{n}\geq\displaystyle\int_{m+1}^{n}\dfrac{1}{t}dt=\log n-\log(m+1)$, in particular, $x_{2m+2}-x_{m}=\log 2$ does not converge to zero as $m\rightarrow\infty$.

Answer 2

I)-Take $$S_n =\sum_{k=1}^n \frac1k \to \infty$$ That is $S_n$ is not Cauchy.

But $$ S_{n+1}-S_n =\frac1n\to 0$$

II)- Also take $$u_n = \sqrt n\to \infty$$That is $u_n$ is not Cauchy. But $$ u_{n+1}-u_n =\sqrt{n+1}-\sqrt{n} =\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$$

III)- $$v_n=\ln n\to \infty$$

and $$ v_{n+1}-v_n =\ln({n+1})-\ln{n} =\ln\left(\frac{n+1}{n}\right)\to 0$$

Answer 3

You are probably having a conceptual block in that the two terms probably seem intuitively equivalent. But they aren't.

1) $x_{n+1} - x_{n} \to 0$ means that each pair of consecutive terms get close together. But all the terms in general do not get close together as a whole. It's still possible that terms that are not consecutive can get very far apart.

A good example of this would be $x_n = \log n$. Then $\log (n+1) - \log n = \log \frac {n+1}n \to \log 1 = 0$. The terms get "close together". But on the whole the terms are completely unbounded. $\log n - \log m \to ?????$. $x_{n+1}$ and $x_n$ must be close togeter. But $x_{n}$ and $x_{n^{500gazillionbillion}}$ do not need to be close together at all.

2) $x_n$ is Cauchy means that to terms get "close together" "as a whole". It's not just that the consecutive terms $x_{n+1}$ and $x_n$ get close together. It's that all terms $x_n$ and $x_m$ get close together. As $M$ gets larger then $x_n$ and $x_m$ for $m,n > M$ must get close together. $x_n$ and $x_{n^{500gazillionbillion}}$ do have to be close together.

Of course that is informal.

The formal definition of $1$ is:

1 or b)) For any $\epsilon > 0$ there is an $M_{\epsilon}$ so that for all $n > M_{\epsilon}$ then $|x_n - x_{n+1}| < \epsilon$. The "english translation" of that is "we can get any two consecutive terms as close together as we like" (as close together as $\epsilon$) "by taking terms far enough down in the series" (past the $M_{\epsilon}$th term).

2 or a)) "cauchy" means for any $\epsilon > 0$ we can find an $M_{\epsilon}$ so that for all $n, m \ge M_{\epsilon}$ that $|x_n - x_m| < \epsilon$. The "english translation" of that is "we can get any two terms not just the consecutive terms $x_n$ and $x_{n+1}$ but any two terms such as $x_n$ and $x_{n^{500gazillionbillion}}$ as close together as we like by taking terms far enough down in the series".

Obviously 2) implies 1). If any two terms $n, m> M_{\epsilon}$ imply that $|x_n - x_m| < \epsilon$, it we work with two consecutive terms $n, n+1 >M_{\epsilon}$.

But the reverse is not true.

Using my example of $x_n = \log n$, then for any $\epsilon > 0$ and any $n > M$, I can find an $m > n > M$ so that $\log m - \log n = \log \frac mn > \epsilon$ simply by taking $\frac mn > 10^\epsilon$. Which is always possible no matter how large I let $n$ be. I just have to let $m$ be much larger.

......

For some reason, analysis classes don't use my logarithm example[$*$] (probably because an analysis class likes to break things down and rebuild things and doesn't want to use "common scense" knowledge of logariths which they will have to prove later... after teaching the concept of cauchy.. Instead they nearly always use the Harmony Series. $\frac 1 + \frac 12 + \frac 13 + \frac 14 + .....\to \infty$ despite the $\frac 1n - \frac 1{n+1} \to 0$.

Google "Harmonic series" for an explanation why $\frac 1 + \frac 12 + \frac 13 + \frac 14 + .....\to \infty$

This means that for any $M$ we can find a $w$ so that $\frac 1 + \frac 12 + ........ \frac 1w > M$ and for any $\epsilon > 0$ and $n$ we can find an $m$ so that $\frac 1 + \frac 12 + ..... + \frac 1m > \epsilon + (1 + \frac 12 + ..... + \frac 1n$.

Or in other words $x_m - x_n = (\frac 1 + \frac 12 + ..... + \frac 1m) -(\frac 1 + \frac 12 + ..... + \frac 1n) > \epsilon$.

So $\{x_n\}=\{(1 + \frac 12 + .... + \frac 1n)\}$ is not Cauchy.

====

[$*$]Actually the harmonic series and the natural logarithm are closely related. $\le x = \int_{1}^x \frac 1x dx\approx \frac 12 + \frac 13 + \frac 14 + .... + \frac 1n$ if $x \approx n$ and $n $ is large enough that taking $dx = \delta x \approx 1$ is reasonable approximation (which it is if $x$ is very large).

Which gives us a significant hint in proving the harmonic series converges. $\ln m - \ln n = \ln \frac mn$ which can be arbitrarily large.

So $( \frac 12 + ... + \frac 1m) - (\frac 12 + ... + \frac 1n) = \frac 1n +\frac 1{n+1} + \frac 1{n+2 } + ..... \frac 1m > \frac 1n + \frac 1n + \frac 1n + .... + \frac 1n = \frac {m-n}n = \frac mn - 1$ whih can be arbitrarily large.

Ex. $\frac 12 + \frac 13 + \frac 14 .... >$

$\frac 12 + (\frac 13 + \frac 14 ) + (\frac 15 + \frac 16 + \frac 17 + \frac 18) + ... + (\frac 1{2^n + 1} + .... + +\frac 1{2^{n+1}} ) + .... >$

$\frac 12 + (\frac 14 + \frac 14) + (\frac 18 + \frac 18 + \frac 18 + \frac 18) + (\frac 1{16} + .... \frac 1{16} + ..... + (\frac 1{2^{n+1}} + ..... + \frac 1{2^{n+1}}) + .... =$

$\frac 12 + \frac 24 + \frac 48 + ..... + \frac {2^n}{2n^{n+1}} + ..... =$

$\frac 12 + \frac 12 + \frac 12 + ....... $

which is unlimited.