Prove that $\frac{n}{2^n}$ is a null sequence from $\epsilon$ definition of limit I am trying to prove that $\frac{n}{2^n}$ is a null sequence using the $\epsilon$ definition of a limit. Now I chose to use the fact that $2^n  > n^2$ for $n > 4$. I said let $\epsilon > 0$ be given. Then for $n > 4$, $\vert\frac{n}{2^n}\vert < \vert\frac{n}{n^2}\vert = \vert\frac{1}{n}\vert.$ The value of $N$ that I chose in the definition of convergence was $[\frac{1}{\epsilon}]+4$. Would this value of $N$ work in this case? I obtained $[\frac{1}{\epsilon}]+1$ from rearranging $\vert\frac{1}{n}\vert < \epsilon$, but of course this would not necessarily guarantee that $n > 4$, which is what we need to produce the first inequality, hence why I added $3$ to this value. 

A null sequence
eventually gets as small as you want.
Initial terms do not matter.
So you can start at $n=4$
or
$n=10000$
or
$n=10^{10^{10}}$.
All that matters is that,
for any $c > 0$,
there is a $N(c)$
such that
$|a_n| < c$
for $n > N(c)$.
Once you have shown that
there is a $m$ such that
$2^n > n^2$
for all $n > m$,
then,
since
$\dfrac{n}{2^n}
\lt \dfrac1{n}$
for $n > m$,
the rest is easy.
