# Math Help - Another divergence proof: Prove 1/a_n converges to ∞ (i.e. 1/a_n diverges).

1. ## Another divergence proof: Prove 1/a_n converges to ∞ (i.e. 1/a_n diverges).

If $a_n$ > 0 $\forall$ n $\in$ $\mathbb{N}$ and $\lim_{n\rightarrow\infty}$ $a_n$ = 0, prove that $\dfrac{1}{a_n}$ $\rightarrow$ $\infty$.

I haven't been able to write up a proof for this one yet. I have a few ideas down:

Since { $a_n$} $\rightarrow$ 0, by definition of convergence, I know | $a_n$| < $\varepsilon$, or
- $\varepsilon$< $a_n$< $\varepsilon$.

Starting ideas for proof:
Assume $\lim_{n\rightarrow\infty}$ $a_n$ = 0 and $a_n$ > 0 $\forall$ n $\in$ $\mathbb{N}$.
{ $a_n$} $\rightarrow$ 0 $\Rightarrow$| $a_n$ - 0| < $\varepsilon$ $\Rightarrow$ | $a_n$| < $\varepsilon$ $\Rightarrow$ | $\dfrac{1}{a_n}$| > $\dfrac{1}{\varepsilon}$, since $\dfrac{1}{\varepsilon}$ > 0 .....

I was working with my classmates, and we were guessing that we would reach a contradiction somehow using the definition of convergence, and then by ______ (we don't know what exatly) we will conclude that the sequence has to diverge by contradiction to our assumption.

Any help, suggestions, tips, corrections, guidance, etc. is greatly appreciated. Thank you very much for looking and for your time!

2. Hello,

Just use the definitions

$a_n \to 0 ~:$

$\forall \varepsilon >0,~ \exists N \in \mathbb{N},~ \forall n \geq N,~ |a_n|=a_n< \varepsilon$

The inequality can be rewritten :
$\frac{1}{a_n} > \frac 1 \varepsilon$

So finally, we have :
$\forall A>0,~ \exists N \in \mathbb{N},~ \forall n \geq N,~ \frac{1}{a_n} > A$
where in fact $A=\frac 1 \varepsilon$, but it doesn't matter since we take random $\varepsilon$

And this is the exact definition of $\frac{1}{a_n} \to +\infty$

If you can't visualize it, take $b_n=\frac{1}{a_n}$ and rewrite the definition of "limit goes to + infinity" for b_n.

3. Originally Posted by Moo
Hello,

Just use the definitions
Awesome. Thanks!