Can you please verify if my solution is correct. let $(a_n=\frac{1}{n})$ be a sequence of real numbers. show that $\sup(a_n)=1$ and $\inf(a_n)=0$.
since $$n+1>n$$ $$\frac{1}{n+1}<\frac{1}{n}$$ $$a_{n+1}<a_n$$ therefore, $(a_n)$ is a decreasing function hence, $\sup(a_n)=a_1=1$ and $\inf(a_n)=\lim\limits_{n \to \infty} a_n = 0$.
I already have proved the limit to be $0$.
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$\begingroup$To show that $\inf_n (\frac{1}{n}) = 0$ you need to show two things:
i) $0$ is a lower bound. This is easy since $\frac{1}{n} > 0$ for all $n$.
ii) $0$ is the greatest lower bound. So suppose there exists $\beta$ a lower bound of $\frac{1}{n}$. Assume, by way of contradiction, that $\beta > 0$. Now there exists an $N \in \mathbb{N}$ such that $\frac{1}{N} < \beta$, contradicting the fact that $\beta$ is a lower bound.
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