Hard time understanding Cauchy criterion and the convergence criterion using a Cauchy series as an example

Question

$$b_n=\frac{x}{1}+\frac{x+1}{3}+\frac{x+2}{5}+\frac{x+3}{7}...+\frac{x+n}{2n+1}$$ $$n>1$$

I have this series and I have no idea how to solve it.I have seen the formula but I don't understand it completely:$$|a_n-l|\leqε$$ I guess that module | | represents the fact that every constant that has a negative value is changed into a positive value???Is that what it means? So,if $a_n$ is a function and is far away from the position of the limit L and is negative that means the values plugged into the module are changed into positive? I looked at the Cauchy definition of the series and it showed me the same module but with different terms $|a_{n+1}+a_{n+2}+a_{n+3}...+a_{n+p}|<ε$.

Answer 1

You have the infinite series $$b_n=\frac{x}{1}+\frac{x+1}{3}+\frac{x+2}{5}+\frac{x+3}{7}...+\frac{x+n}{2n+1}$$

You are looking for an answer about the convergence of your series.

There is a divergence test which applies to your series

The series $$\sum _1^\infty a_n$$ diverges if the term $a_n$ does not tend to zero as $n$ goes to infinity.

Since $$\lim _{n\to \infty} \frac {x+n}{2n+1} = \frac {1}{2} \ne 0$$

Your series diverges. The Cauchy Criterion does not apply here for this series.

Answer 2

$|A|$ denotes the absolute value of $A$. It does not change $A$ into $|A|$. Numbers do not change into other numbers. If you write $-3$ and then you put $|\,|$ around it then you have only changed what you have written but you have not changed the number $-3$ into the number $3.$

Definition.

$|A|=A$ if $A\ge 0.$

$|A|=-A$ if $|A|<0.$

It helps to imagine this geometrically, with numbers represented as locations on a line, with $|A|$ as the distance from $A$ to $0.$ And $|x-y|$ is the distance from $x$ to $y$, which is the same as the distance from $y$ to $x$, which is also the distance from $0$ to the number $y-x.$

In this subject there are short forms and figures of speech that are often used among those who understand what the literal meanings are, but to learn it you must adhere to the literal meanings. Which is sometimes disappointing because sometimes there's not much to it, as in the above definition of $|A|$. It means what it says, and no more.

Answer 3

I think that the OP is mixing up the definition of convergence with the Cauchy criterion. I will write a description of both.

Convergence

Definition: The sequence $x_n$ converges to $X$ when this holds: for any $\epsilon > 0$ there exists $K$ such that $|x_n − X| < \epsilon$ for all $n > K$.

Informally, this says that as $n$ gets larger and larger the numbers $x_n$ get closer and closer to $X$.

Cauchy's criterion

Definition: The sequence $x_n$ converges to something if and only if this holds: for every $\epsilon > 0$ there exists $K$ such that $|x_n − x_m| < \epsilon$ whenever $n, m > K$.

Informally, this says that as $n,m$ get larger and larger the numbers $x_n$ and $x_m$ get closer and closer to each other.

In terms of your example, I tried to show that the sequence was Cauchy by writing

$$x_n = \frac{x+n}{2n+1}$$ $$x_m = \frac{x+m}{2m+1}$$

then for $n,m$ large enough we have

\begin{align}|x_n - x_m| &= \Big|\frac{x+n}{2n+1} - \frac{x+m}{2m+1}\Big| \\&= \Big|\frac{(x+n)(2m+1) - (x+m)(2n+1)}{(2n+1)(2m+1)}\Big| \\&= \Big|\frac{2mx - m - 2nx + n}{(2n+1)(2m+1)}\Big| \\&\leq \Big|\frac{2mx+ n }{(2n+1)(2m+1)}\Big| \\&\leq \Big|\frac{2mx}{(2n+1)(2m+1)}\Big| + \Big|\frac{n}{(2n+1)(2m+1)}\Big|\\&\leq \Big|\frac{2mx}{4mn}\Big| + \Big|\frac{n}{4mn}\Big|\\&= \Big|\frac{x}{2n}\Big|+\Big|\frac{1}{4m}\Big|\\& \leq \Big|\frac{x}{2}\Big|+\Big|\frac{1}{4m}\Big| \end{align}

Since $x$ can be made arbitrarily large, the Cauchy criterion isn't satisfied (we cannot show that $|x_n - x_m| < \epsilon$).