Limit of a sequence of real numbers 2 I want to show using the definition of limit of a sequence that
$$\lim_{n\to\infty}{ne^{-n}}=0.$$
Here $0\notin\mathbb{N}$. This is my attemp:
We note that
\begin{align*}
\lim_{n\to\infty}{ne^{-n}}=0 \iff \forall\epsilon>0 :\exists N\in\mathbb{N} : \forall n\geq N : |ne^{-n}-0|<\epsilon.
\end{align*}
and
\begin{align*}
&|ne^{-n}-0|<\epsilon\\
&\iff |ne^{-n}|<\epsilon\\
&\iff \left|\frac{n}{e^n}\right|<\epsilon
\end{align*}
since $n$ must be more than a natural number, we have that $n/e^n>0$, so
\begin{align*}
&\frac{n}{e^n}<\epsilon\\
&\implies \ln\left(\frac{n}{e^n}\right)<\epsilon\\
&\implies \ln(n)-\ln(e^n)<\ln(\epsilon)\\
&\implies \ln(n)-n<\ln(\epsilon)\leq \epsilon -1\\
&\implies \frac{n-1}{n}<\ln(n)<n+\epsilon-1\\
&\implies \frac{n-1}{n}<n+\epsilon-1\\
&\implies n-1<n^2+n\epsilon-n\\
&\implies n^2+n\epsilon-n-n+1>0\\
&\implies n^2+n(\epsilon-2)+1>0 \Longleftrightarrow:\varphi
\end{align*}
I can´t see how I can get something useful from $\varphi$. Can anyones help to find the way please?
Thank you so much.
 A: I like your initiative, but there's a serious and fundamental problem with this attempt: the direction of the logic. Remember, you need $n / e^n < \varepsilon$ to be the conclusion of your logic, not the premise. It's fine to work from this proposition, but if you do, each step should logically imply the previous step, not the next step. For example, where you state
$$\ln(n)-\ln(e^n)<\ln(\varepsilon) \implies \ln(n)-n < \varepsilon -1,$$
this is an example where the logic does not flow in the correct direction. While this implication is valid, it is not helpful. We are interested in statements that imply $\ln(n)-\ln(e^n)<\ln(\varepsilon)$, not its consequences.
If you want information on how to tackle this well, you can check out this question. I'm sure there are others on this site too. Often, this is done with squeeze theorem. Other approaches, using heavier machinery, involve l'Hopital's rule (showing the stronger result that $\lim_{x \to \infty} xe^{-x} = 0$, where $x \in \Bbb{R}$), or by proving $\sum ne^{-n}$ is a convergent series using one of the many tests, and using the fact that terms of convergent series must tend to $0$.
A: No, this is too complicated. There is a simple estimation as follows. First,
$$
2^n=(1+1)^n>\dfrac {n(n-1)}{2}
$$
For any $\epsilon>0$, let $N=\left[\dfrac2{\epsilon}+1\right]$. Then for any $n>N$, there is
$$
\left|\dfrac{n}{e^n}-0\right|<\dfrac{n}{2^n}<\dfrac2{n-1}<\epsilon
$$
