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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?

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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 $\dfrac1{Nt}<\varepsilon$.

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

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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 $n \ge N$:

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

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