Why doesn't $d(x_n,x_{n+1})\rightarrow 0$ as $n\rightarrow\infty$ imply ${x_n}$ is Cauchy? What is an example of a sequence in $\mathbb R$ with this property that is not Cauchy?
I know that Cauchy condition means that for each $\varepsilon>0$ there exists $N$ such that $d(x_p,x_q)<\varepsilon$ whenever $p,q>N$.
 A: $$\textstyle 0,\,{1\over 2},\, 1,\,  {2\over 3},\,{1\over 3},\,{0},\,{1\over 4},\,{2\over 4},\,{3\over 4},\,1,\,{4\over 5},\,{3\over 5},\,{2\over 5},\,{1\over 5},  \,0, \ldots$$
A: Another example in the line with the usual metric is $x_n = \sqrt{n}$.
A: The standard example is the sequence $(s_n)$ of partial sums of the harmonic series. Formally,
$$s_n=\sum_{k=1}^n \frac{1}{k}.$$  
Note that $d(s_n,s_{n+1})=\frac{1}{n+1}$. It is clear that $d(s_n, s_{n+1})\to 0$ as $n\to\infty$.
But the sequence $(s_n)$ is not Cauchy. For given any $m$, we can find $n$ such that $d(s_m,s_n)$ is arbitrarily large. This is because the sequence $(s_n)$ diverges to infinity.  We omit the proof, since you have likely already seen a proof that $\sum_{k=1}^\infty \frac{1}{k}$ diverges.
A: Let 
$$
s_{n}:= \sum_{i=1}^{n} \frac{1}{i}.
$$
Then 
$$
s_{n+1}-s_{n} = \frac{1}{n+1} \to 0
$$
as $n\to \infty$. 
However, this sequence is not Cauchy, as for any $\varepsilon > 0$, there is no $N$ such that 
$
|s_{m}-s_{n}| < \varepsilon
$
for all $n,m\ge N$. In fact, 
$$
\lim_{m\to\infty} |s_{m}-s_{n}| = \infty,
$$
for any $n$.
The "for all" quantifier on $m$ and $n$ is what confuses people about Cauchy sequences. 
A: Let the metric space be $\mathbb R$ with respect to the usual Euclidean distance metric $|.|$ , consider the 
sequence $x_n:=\log n , \forall n \in \mathbb Z^+ $ , then $d(x_{n+1}, x_n)=|\log(n+1)-\log n|=|\log\big(1+\dfrac1n\big)| \to \log 1=0$ as $n \to \infty$ , but off-course
$(\log n)_{n=1}^\infty$ is not a bounded sequence , hence is not Cauchy 
