Why $x_n:=\sqrt{n}$ is not a Cauchy sequence, but $\lim|x_{n+1}-x_n|=0$ Since $\lim|x_{n+1}-x_n|=0$, it seems that every two adjacent items are close to each other, but why accumulatively, the sequence is divergent?
 A: A sequence $(x_n)$ is Cauchy iff for each $\epsilon > 0$ there exists some $N$ such that $|x_n - x_m| < \epsilon$ for $n, m > N$, not merely for $m = n + 1$. In fact, the difference $x_n - x_{n-1} \approx \frac{1}{2\sqrt{n}}$; since $\sum_n \frac{1}{\sqrt{n}}$ diverges, the difference $x_n - x_m = (x_n - x_{n-1}) - (x_{n-1} - x_{n-2}) + \cdots + (x_{m+1} - x_m)$ becomes arbitrarily large for $n\gg m$.
A: An intuitive complement to anomaly's answer:  The statement that $\lim |x_{n+1}-x_n|=0$ says neighboring terms get close to each other, but they could get close too slowly so the sum of the remaining differences diverges.  Your example is a good one.  No matter what $N$ you give me, I can find an $n$ so that $\sqrt n \gt N$.  The difference of successive terms is about $\frac 1{\sqrt n}$ but the sum of a lot of those gets huge.  For a sum to converge, you don't just need the individual terms to get small, you need them to get small fast enough and $\frac 1{\sqrt n}$ is not fast enough.
A: By definition a sequence is Cauchy if for every $m$ and $n$, the relation $|x_m-x_n|<\varepsilon$ holds. With $m=k^2$ and $n=(k+1)^2$ you can make a subsequence which doesn't satisfy $|x_m-x_n|<\frac12$.
A: Treat this as a infinite sum, $\lim x_n=(x_2-x_1)+(x_3-x_2)+\cdots+(x_n-x_{n-1})+\cdots$ and $x_n-x_{n-1}\rightarrow 0$.
However, a sum of infinite many infinitesimals can be infinity. 
There are many examples, $1+1/2+\cdots +1/n+\cdots\rightarrow +\infty$
