Prove that $\lim(x_n)=0$ using definition of limit of sequences. Let $x_n:=\dfrac{1}{\ln(n+1)}\space\forall \space n\in N$
Use the definition of limit of sequences to prove that $\lim(x_n)=0$
I tried to use $e^n>n+1\Rightarrow n>\ln(n+1)$ but that gives me $\dfrac{1}{n}<\dfrac{1}{ln(n+1)}$ which doesn't seem of much use.
Please help.
 A: To show that a sequence $(x_n)$ converges to $0$, you to prove that for every $\epsilon > 0$, there exists a natural $N$ such that for every natural number $n > N$, you have $\vert x_n\vert < \epsilon$.
So let's take $\epsilon > 0$ arbitrarily small. We have
$$
\dfrac{1}{\ln(n+1)} < \epsilon
$$
if and only if
$$
n+1 > e^{1/\epsilon}.
$$
So you can fix some natural number $N$ larger than $e^{1/\epsilon}$ and you will have for every natural number $n > N$ that (using monotonicity of log and positivity of the sequence)
$$
0 < x_n < x_N < \epsilon,
$$
which gives the result.
A: Hint: show that for any real number $\epsilon>0$ you can find an integer $N_{\epsilon}>0$ such that $x_n<\epsilon\;\forall n > N_{\epsilon}$
A: Here's one way to do it.
First, a word on notation.
Let $\mathbb{N}$ refer to the positive integers.
Let $\mathbb{N}_>$ refer to the upward closed sets of $\mathbb{N}$.
$$ \mathbb{N}_> \stackrel{\text{def}}{=\!=} \{\{1, 2, 3, 4, \cdots\}, \{2, 3, 4, 5 \cdots\}, \{3, 4, 5, 6 \cdots\}, \cdots \} $$
Let a variable with a dot on top of it, like $\dot{w}$, refer to a non-empty set of real numbers. This notation is nonstandard; I'm using it to reduce the number of quantifiers.
Applying a function to a set variable $f(\dot{w})$ produces a set. If a set appears as an argument to to $<$, $\le$, ..., then the statement is true if and only if the comparison applies to every element of the set.
Here's the definition of $x_n$ given in the question
$$ x_n \stackrel{\text{def}}{=\!=} \frac{1}{\ln(n+1)} \;\; \text{for all $n$ in $\mathbb{Z}_{\ge 0}$ } $$
We are asked to show the following. Note that $n$ only ranges over non-negative integers.
$$ \lim_{n \to \infty} x_n = 0 $$
This is equivalent to the following limit if we constrain $n$ to range over positive integers.
$$ \lim_{n \to \infty} \frac{1}{\ln(n)} = 0 $$
In order to prove this, we start with the following fact:
$$ \forall \sigma \in \mathbb{R}_{> 0} \mathop. \exists \dot{w} \in \mathbb{N}_{>} \mathop. \;\;\;\; \dot{w} > \exp(\sigma) $$
In order to prove it, we pick, $1 + 3^{\lceil \sigma \rceil}$ as the minimum value of $\dot{w}$.
Since $\exp$ is monotonic increasing, its inverse $\ln$ is also monotonic increasing.
$$ \begin{align} \ln(\dot{w}) & > \ln \circ \exp(\sigma) \\\\ \ln(\dot{w}) & > \sigma \end{align} $$
Let's apply the substitution $\delta = \frac{1}{\sigma}$
$$ \forall \delta \in \mathbb{R}_{> 0} \mathop. \exists \dot{w} \in \mathbb{N}_{>} \mathop. \;\;\;\; \ln(\dot{w}) > \frac{1}{\delta} $$
And take the reciprocal of both sides.
$$ \frac{1}{\ln(\dot{w})} < \delta $$
Note that $\ln$, when applied to an element of $\mathbb{N}$ is always non-negative. Therefore, we can take the absolute value of the left hand side.
$$ \left| \frac{1}{\ln(\dot{w})} \right| < \delta $$
Thus, as desired
$$ \lim_{n \to \infty} \frac{1}{\ln(n)} = 0 \;\;\; \text{where $n$ is in $\mathbb{N}$} $$
