I want to show that the infimum of the set containing the terms of the harmonic sequence is 0. Can I simply argue that because the harmonic sequence converges to 0 then the infimum of the set containing terms of the harmonic sequence is 0?
Our recursive definition of a sequence is $1/n$ where $n$ starts at n=1 and n goes to infinity.
Existence of Sequence in Set of Real Numbers whose Limit is Infimum?
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In this particular case, your intuition is correct, but you have to rule out a couple of exceptions.
Let $a_n=\frac{1}{n}$. First of all, the infimum $m$ exists, because the set $\{a_n : n\in \mathbb{N} \}$ is bounded below. In fact, since $a_n>0$, the infimum is $m\ge 0$. Now assume for the sake of contradiction that $m>0$. Since $\lim_{n\to\infty} a_n=0$, there exists $N$ such that $a_n<m$ for $n>N$, which contradicts $m$ being the infimum. So $m=0$.
In summary, you need convergence and $a_n>0$, or another equivalent requirement.
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How do you show the limit is $0$ without showing the infimum is $0\text{?}$
How about this: If $\varepsilon>0$ is a lower bound of $\left\{ 1, \dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \ldots \right\}$ then $1/\varepsilon$ is an upper bound of $\{1,2,3,4,\ldots\},$ so there must be a least upper bound $M$ of $\{1,2,3,4,\ldots\}$. That means $M-1$ is not an upper bound, so some member $n$ of $\{1,2,3,4,\ldots\}$ is $> M-1.$ But then $n+1$ is also a member of $\{1,2,3,4,\ldots\}$ and is $>M,$ so we have a contradiction.
Not exactly; the convergence alone does not ensure that conclusion. Indeed, there is a converging sequence $(x_{n})$ with limit not $\inf_{n}x_{n}$ (what is an example?). If $x_{n} := 1/n$, then $(x_{n})$ is decreasing ; so it just so happened that $\inf_{n}x_{n} = \lim_{n}x_{n}$ (Try to prove this; could you?).