Question: $(x_n)_{n=1}^\infty$ is a sequence with $x_n\neq0 $ for all $n$, also let $x_n$ tend to infinity. Let $(y_n)_{n=1}^\infty$ be defined by $y_n=\frac{1}{x_n}$, show that this converges to zero.
Definition for tending to infinity:$\forall K \in \mathbb{R} \exists N\in \mathbb{N} \forall n \in \mathbb{N} ,n>N:x_n>K $
Definition for convergence to zero:$\forall \varepsilon > 0 \exists N\in \mathbb{N} \forall n \in \mathbb{N} ,n>N:∣x_n∣<\varepsilon $
Idea: I know that $x_n$ tends to infinity since this is assumed. Then it must satisfy the defintion, therefore there must exist an $N$ which satisfies the defintion which i will call $N_x$ to prevent confusion.
To prove that $y_n$ converges to zero then I should choose $N_y=\lceil\frac{1}{N_x}\rceil$
Proof (attempt): Given $\varepsilon>0$ choose $N_y=\lceil\frac{1}{N_x}\rceil$. Given $n \in \mathbb{N}$, $n>N$ we have $∣y_n∣$$=∣\frac{1}{x_n}∣\leq \frac{1}{N_x}\leq\varepsilon $
Not entirely sure my selection for $N_y$ is correct, I understand why the statement is true however the proving put I am struggling.
Best Answer
Take $\varepsilon>0$. Since $\lim_{n\to\infty}x_n=\infty$, there is some $N\in\mathbb N$ such that $n\geqslant N\implies x_n>\frac1\varepsilon$. But then $n\geqslant N\implies 0<\frac1{x_n}<\varepsilon$. In particular, $\left\lvert\frac1{x_n}\right\rvert<\varepsilon$.