1. **Stating the problem:** We want to show that for a sequence $a_n > 0$ for all $n$, $\lim_{n\to\infty} a_n = \infty$ if and only if $\lim_{n\to\infty} \frac{1}{a_n} = 0$. 2. **Recall definitions:** - $\lim_{n\to\infty} a_n = \infty$ means for every $M > 0$, there exists $N$ such that if $n > N$, then $a_n > M$. - $\lim_{n\to\infty} \frac{1}{a_n} = 0$ means for every $\varepsilon > 0$, there exists $N'$ such that if $n > N'$, then $\left| \frac{1}{a_n} - 0 \right| = \frac{1}{a_n} < \varepsilon$. 3. **Proof of $\Rightarrow$ (If $\lim_{n\to\infty} a_n = \infty$, then $\lim_{n\to\infty} \frac{1}{a_n} = 0$):** - Let $\varepsilon > 0$ be given. - Since $a_n \to \infty$, there exists $N$ such that for all $n > N$, $a_n > \frac{1}{\varepsilon}$. - Taking reciprocals and using $a_n > 0$, for all $n > N$: $$\frac{1}{a_n} < \varepsilon$$ - Therefore, $\lim_{n\to\infty} \frac{1}{a_n} = 0$. 4. **Proof of $\Leftarrow$ (If $\lim_{n\to\infty} \frac{1}{a_n} = 0$, then $\lim_{n\to\infty} a_n = \infty$):** - Let $M > 0$ be given. - Since $\frac{1}{a_n} \to 0$, there exists $N'$ such that for all $n > N'$, $$\frac{1}{a_n} < \frac{1}{M}$$ - Taking reciprocals (with $a_n > 0$), for all $n > N'$: $$a_n > M$$ - Thus, $\lim_{n\to\infty} a_n = \infty$. 5. **Conclusion:** We have shown each direction holds by exploiting the reciprocal relationship preserving inequalities for positive sequences. Hence, $$\lim_{n\to\infty} a_n = \infty \iff \lim_{n\to\infty} \frac{1}{a_n} = 0$$