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I'm given the following problem in my analysis homework set.

A sequence $a_n$ of positive real numbers has the following property: for every $L \in \mathbb N$ there is a natural number $C$ such that $a_n > L$ whenever $n>C$. Prove that the sequence $\frac{1}{a_n}$ converges to zero.

I'm not sure how to even begin thinking about this problem with the information given. I guess I'm just not sure how to interpret what it's saying. For information I'm able to use, I know the standard limit laws and have proved that $\frac{1}{n}$ converges to zero.

Any guidance would be appreciated, thank you.

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    $\begingroup$ You need to show that for any $\epsilon >0$ there exists a $C\in\mathbb N$ such that for all $n\in\mathbb N$ satisfying $n\ge C$, we have $$\left|{1\over a_n} -0 \right| = \left|{1\over a_n}\right|< \epsilon.$$ $\endgroup$
    – Decaf-Math
    Feb 22, 2017 at 6:35

1 Answer 1

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To prove that $1/a_n$ converges to zero, you have to show that for any $\epsilon>0$, there exists $C$ such that $1/a_n<\epsilon$ whenever $n>C$. In other words, you want that $a_n>1/\epsilon$ whenever $n>C$. Now, what $L$ (in terms of $\epsilon$) should you take so that this property is satisfied?

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