Question

Show directly from the definition that Xn= (2n+1)/(n) is a cauchy sequence

Answer 1

To show the sequence is Cauchy, you must show that there is some \(m,n\ge N\) for which \(|x_n-x_m|<\epsilon\) for \(\epsilon>0\) and \(m\le n\).

In this particular case, you must show
\[\left|\frac{2n+1}{n}-\frac{2m+1}{m}\right|<\epsilon\]
As part of your preliminary analysis - the work for determining a proper \(N\) - you assume the above inequality and try to derive a lower bound.
\[\begin{align*}\left|\frac{2n+1}{n}-\frac{2m+1}{m}\right|&=\left|\frac{2mn+m}{mn}-\frac{2mn+n}{mn}\right|\\
&=\left|\frac{2mn+m-2mn-n}{mn}\right|\\
&=\left|\frac{m-n}{mn}\right|\\
&\le\left|\frac{m}{mn}\right|-\left|\frac{n}{mn}\right|\\
&=\left|\frac{1}{n}\right|-\left|\frac{1}{m}\right|\\
&\le\frac{1}{n}\\
&\le\frac{1}{N}
\end{align*}\]
From this, you would use \(\dfrac{1}{N}<\epsilon\), which means, working backwards, you would use \(N=\dfrac{1}{\epsilon}\).

Answer 2

Ah I understand! So cauchy sequence and cauchy series are proven differently. If i had a series i would have to show that \[|x _{n}+x _{n+1}+...x _{m}|\] right?

Answer 3

|xn+xn+1+...xm| < ϵ *

Answer 4

For a series, you would show that the partial sums get closer and closer together:
\[\left|\sum_{k=1}^nx_k-\sum_{k=1}^mx_k\right|<\epsilon\]
These proofs tend to involve more uses of the triangle inequality, if I'm remembering correctly.

Answer 5

If you use \(n=m+p\) for some \(p\), you would then break up the terms:
\[\begin{align*}\left|\sum_{k=1}^nx_k-\sum_{k=1}^mx_k\right|&=\left|\left(x_1+\cdots+x_n\right)-\left(x_1+\cdots+x_m\right)\right|\\
&=\left|\left(x_1+\cdots+x_m+\cdots+x_{m+p}\right)-\left(x_1+\cdots+x_m\right)\right|\\
&=\left|x_{m+1}+\cdots+x_{m+p}\right|\\
&\le|x_{m+1}|+\cdots+|x_{m+p}|
\end{align*}\]