I'm a student doing a real analysis course and I'm working through my problem sheet and I'm just hoping for some feedback on the following question, just to make sure I'm on the right track with my thinking. The question is follows...
Suppose that $\{y_n\}$ is a sequence of positive real numbers. Then $y_n \rightarrow \infty$ as $n \rightarrow \infty$ means for every $M \in \mathbb{R}$ there exists $N \in \mathbb{N}$ such that.
$$n \geq N \implies y_n > M$$ suppose that $y_n \rightarrow \infty$ as $n \rightarrow \infty$. Prove that $\frac{1}{y_n} \rightarrow 0$ as $n \rightarrow \infty$
I take the definition given by the question.
$$n \geq N \implies y_n>M$$
$$n \geq N \implies \frac{1}{y_n}<\frac{1}{M}$$
$$n \geq N \implies \frac{1}{y_n}<\epsilon$$
$$n \geq N \implies \bigg| \frac{1}{y_n}\bigg|<\epsilon$$
As the question says $y_n$ is a postive sequence of real numbers.
$$n \geq N \implies \bigg| \frac{1}{y_n}-0\bigg|<\epsilon$$
Thus it is in the correct form impling that $\frac{1}{y_n}$ coverges to zero.
Thanks for your time!
At first sight it may be fine, but the definition states that you shall begin with $\varepsilon >0$ and then find a $N \in \mathbb N$ such that you get your desired result. This is where $M$ and the definition by the question comes into play, where you shall use $M$ to express a relation with $\varepsilon$. Your intuition is on the right way, but be careful to form the definition via your expressions in a correct and rigorous way.