Theorem: For any sequence , $b_n >0$ then $b_n \to 0$ iff limit of $\frac{1}{b_n} = \infty$ as $n \to \infty$.
I need help with this problem proof! Thank you very much!
Theorem: For any sequence , $b_n >0$ then $b_n \to 0$ iff limit of $\frac{1}{b_n} = \infty$ as $n \to \infty$.
I need help with this problem proof! Thank you very much!
The key idea to this problem is that since $b_n > 0$ we have no problem inverting them to $\frac{1}{b_n}$ and if $b_n$ is very small, then $\frac{1}{b_n}$ must be very large.
$(\Rightarrow)$ Suppose that $b_n \rightarrow 0$. Let $K>0$ be large. Then we can choose $N$ large so that $b_N \leq \frac{1}{K}$ and since $b_N > 0$ this means that $\frac{1}{b_N} \geq K$. Thus we see that $\frac{1}{B_N} \rightarrow \infty$.
$(\Leftarrow)$ Suppose that $\frac{1}{b_n}\rightarrow \infty$. Let $\epsilon > 0$ be small. Then we can choose $N$ large so that $\frac{1}{b_N} \geq \frac{1}{\epsilon}$. But then this means that $b_N \leq \epsilon$. Thus we see that $b_n \rightarrow 0$.
If you have tried anything, then at least write down the definitions. Assuming $b_n > 0$, then $b_n \to 0$ is equivalent to $ \forall \varepsilon > 0, \quad \exists N \quad : \forall n > N, \quad b_n = |b_n| < \varepsilon $ and $\frac 1{b_n} \to \infty$ is equivalent to $ \forall M > 0, \quad \exists N \quad : \forall n > N, \quad \frac 1{b_n} = \left| \frac 1{b_n} \right| > M. $ Can you see how those statements are equivalent? The key lies there.
Hope that helps,