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

**ilikedmath** · November 7th 2008, 05:41 PM

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!

**Moo** · November 7th 2008, 11:38 PM

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.

**ilikedmath** · November 8th 2008, 06:58 AM

Originally Posted by Moo

Hello,

Just use the definitions

Awesome. Thanks!

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

LinkBack

Thread Tools

Display

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

Similar Math Help Forum Discussions

Prove if series converges or diverges

Converges or Diverges

Converges, Diverges, Can't Tell

converges or diverges

diverges or converges?

Search Tags