# Proof verification of $\lim_{n \to \infty}\frac{q^n}{n} = 0$ for $|q| < 1$ using $\epsilon$ definition

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

Using the definition of a limit:

$$\lim_{n\to \infty}\frac{q^n}{n} = 0 \stackrel{\text{def}}{\iff} \{ \forall\epsilon>0 ,\exists N\in\mathbb N, \forall n > N : \left|\frac{q^n}{n} - 0\right| < \epsilon \}$$

Consider the following:

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

Redefine $$|q|^n$$: $$|q|^n = \frac{1}{(1+t)^n} \le\frac{1}{(1+nt)}$$ 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 $$\dfrac1{Nt}<\varepsilon$$.

• That's a nice catch, thank you! – roman Nov 21 '18 at 16:50

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 $$n \ge N$$:

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