Real Analysis: Example related to irrational number and rational limit and Cauchy A) a sequence $(x_n)$ of irrational numbers that converges to a rational number
Are the following answer correct?
$x_n = e^1$ 
or
$\sqrt{2}/n$
Give a little explanation if you could
B) a sequence $(x_n)$ that is not Cauchy, but for which $|x_{n+1} - x_n|$ converges to $0$
please give an example
 A: Your answer $a_n = \sqrt{2}/n$ is valid for the first part.  I'm not sure what you meant to type when you put $a_n = e^1$, but as written it is not valid.
This answer to B) is one of those "classic" counterexamples that are helpful, in general, to remember.
$$
x_n = \sum_{k=1}^n\frac 1k
$$
Can you show that $|x_{n+1}-x_n|\to 0$? Can you show that $x_n$ is not Cauchy?
A: A) The sequence $x_{n}=e^{1}$ is a constant sequence that converges to $e$, which is not rational, so this is not such an example.
The sequence $x_{n}=\frac{\sqrt{2}}{n}$ converges to $0$, because for any $\varepsilon>0$ we can pick $N$ large enough (in particular, $N=\frac{2}{\varepsilon}$ works) such that $|x_{n}-0|<\epsilon$ whenever $n>N$. Furthermore, every element in this sequence is irrational, because $\sqrt{2}$ is irrational and the product of a rational number ($\frac{1}{n}$) and an irrational number ($\sqrt{2}) is irrational. Thus this is a sequence of irrationals that converges to a rational.
B) Consider a sequence of partial sums $x_{n}=\sum_{i=1}^{n}\frac{1}{i}$. Can you tell why?

 The terms $|x_{n+1}-x_{n}|=\frac{1}{n+1}$ converge to $0$ as $n$ goes to infinity, but the harmonic series is known to diverge, so the sequence does not converge.

