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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!

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What have you tried so far? –  Michael Albanese Oct 1 '12 at 4:27
    
It is very mean for a downvoter to do so on a new user. I suggest he removes it. As for the question, when you ask for help, it is recommended you show what you have tried so that we might give a hint instead of a full detailed solution! –  Patrick Da Silva Oct 1 '12 at 4:30

2 Answers 2

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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 $b_n \rightarrow 0$.

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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,

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I find it so much easier writing the definitions this way rather than in regular sentences thanks!!! –  Maximiliano Oct 1 '12 at 18:55
    
Sometimes it's just an $\epsilon$ and $N$ game and it's really not that much of a smart statement to prove things rigorously. I don't think you worked a lot from the hint I gave you and I barely wrote down what $A$ and $B$ means in your $A \Rightarrow B$ statement. –  Patrick Da Silva Oct 2 '12 at 1:27

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