Show that $\forall\varepsilon>0$ $\exists N\in\mathbb{N}$ s.t. $\ln\left(\frac{N+1}{N}\right)<\varepsilon$ My question is: how would one prove that: $\forall\varepsilon>0$ $\exists N\in\mathbb{N}$ s.t. $\ln\left(\frac{N+1}{N}\right)<\varepsilon$.
Is it enough to show that as $N\to\infty$, $\ln\left(\frac{N+1}{N}\right)\to 0$ for the inequality to be true?
Thanks!
 A: Let $\epsilon>0$
$\log{\frac{N+1}{N}}<\epsilon \Longleftrightarrow N+1<Ne^{\epsilon} \Longleftrightarrow N(e^{\epsilon}-1)>1$
So take $$N_0=[\frac{1}{e^{\epsilon}-1}]+1$$
So for all $N \geq N_0$ you will have that $\log{\frac{N+1}{N}}<\epsilon$
Here $[x]$ denotes the integer part of $x$.
A: If you know that $\ln$ is continuous with $\ln(1)=0$, then we can indeed use that
$\lim_{n\to\infty} \ln(\frac{n+1}{n})=\ln(\lim_{n\to\infty} \frac{n+1}{n})=\ln(\lim_{n\to\infty} 1-\frac1n)=\ln(1)=0$
But I doubt that and you are asked to give a formal proof.
So let $\varepsilon >0$ be arbitrary. We have to find $N$. How you formulated the task it is enough to give one $N$ for that this holds.
$\ln(\frac{N+1}{N})<\varepsilon \Leftrightarrow \frac{N+1}{N}<e^\varepsilon\Leftrightarrow N+1<Ne^\varepsilon\Leftrightarrow N-Ne^\varepsilon<-1$
$\Leftrightarrow N(1-e^\varepsilon)<-1\stackrel{\cdot (-1)}{\Leftrightarrow} N(e^\varepsilon-1)>1\Leftrightarrow N>\dfrac{1}{e^\varepsilon-1}$
Since $N$ is supposed to be a natural number, we now take $N=\lceil\dfrac{1}{e^\varepsilon-1}\rceil$.
Where $\lceil\cdot\rceil$ notes the ceiling-function https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Definition_and_properties
A: An alternative proof uses the logic of negation of propositions. The negation of the statement
$$
\forall\varepsilon>0 \exists N\in\mathbb{N}-\{0\} \mbox{ s.t. } \ln\left(\frac{N+1}{N}\right)<\varepsilon
$$
is 
$$
\exists \varepsilon>0 \mbox{ s.t. } \forall N\in\mathbb{N}-\{0\} \mbox{ hold } \ln\left(\frac{N+1}{N}\right)>\varepsilon 
$$
Note that 
\begin{align}
\ln\left(\frac{N+1}{N}\right)>\varepsilon 
&
\Leftrightarrow
e^{\ln\left(\frac{N+1}{N}\right)}>e^\epsilon
\\
&
\Leftrightarrow
\frac{N+1}{N} >e^\epsilon
\end{align}
For any fixed $ \epsilon> 0 $ we have  $e^\epsilon>1$. And for $ N $ big enough we have
$$
e^\epsilon>\frac{N+1}{N}>1. 
$$
This contradiction shows that the negation of the statement is false. And so the original statement is true.
A: You need to use the fundamental inequality satisfied by $\log$ function: $$\log x\leq x-1, \, \forall x\geq 1$$ Then we have $$\log\left(\frac{N+1}{N}\right) \leq\frac{1}{N}, \,\forall N\in\mathbb {N} $$ If we choose $1/N<\varepsilon $ then the desired inequality is satisfied by virtue of the above mentioned inequality. Therefore we can take $N$ to be any integer greater $1/\varepsilon $. In particular $N=[1/\varepsilon] +1$ works as desired.
In general when dealing with calculus/analysis stuff one does not solve inequalities (akin to algebraic equations) but rather one simplifies the inequalities to a great extent by using the properties of functions involved. 
