Ytukyg

Prove
$$
lim_{n to infty}frac{q^n}{n} = 0
$$ for $|q| < 1$ using $epsilon$ definition.

Using the definition of a limit:

$$
lim_{nto infty}frac{q^n}{n} = 0 stackrel{text{def}}{iff} { forallepsilon>0 ,exists Ninmathbb N, forall n > N : left|frac{q^n}{n} - 0right| < epsilon }
$$

Consider the following:

$$
left|frac{q^n}{n}right| < epsilon iff frac{|q|^n}{n} < epsilon
$$

Redefine $|q|^n$:
$$
|q|^n = frac{1}{(1+t)^n} lefrac{1}{(1+nt)}
$$
Thus:
$$
frac{|q|^n}{n} < frac{1}{n(1+nt)} < frac{1}{n^2t} < frac{1}{n} < epsilon
$$

So from this we may find $N$ such that:

$$
frac{1}{n} < frac{1}{N} < epsilon
$$

Thus the limit is $0$.

Is it a correct proof?

Your proof is almost correct, but there is a small problem concerning the inequality $dfrac1{n^2t}<dfrac1n$. This is equivalent to $nt>1$. Why would that be true? All you know about $t$ is that $t>0$. So, you should deal with the inequality $dfrac1{nt}<varepsilon$. That is, choose $N$ such that $dfrac1{Nt}<varepsilon$.

Yes, your proof is fine. You should mention, that the inequality $frac{1}{n^2t }<frac{1}{n}$ does not hold for all $n $, but for almost all $n $.

You have::

$dfrac{|q|^n}{n} < dfrac{1}{n^2t}< dfrac{1}{nt},$ $t>0$.

Let $epsilon$ be given.

Archimedean principle:

There is a $N$, positive interger, s.t.

$N >dfrac{1}{t epsilon}$.

For $n ge N$:

$dfrac{|q|^n}{n} <dfrac{1}{nt} le dfrac{1}{Nt} <epsilon$.