# Proving sequences

1. Dec 6, 2011

### gregy6196

if lim(n→∞)an=0 then prove lim(n→∞)$1/an$=0

how do i do this, i know how to proove it geometrically, but how do you write the proof using ε
and $\delta$

Give a counter example to show that the converse is false.

Last edited: Dec 6, 2011
2. Dec 6, 2011

### Erland

There must be a typo here, for if an -> 0 then obviously 1/an -> infinity (if an>0, otherwise it may oscillate between +infinity and -infinity).

3. Dec 6, 2011

### gregy6196

if lim(n→∞) an = ∞ then prove lim(n→∞) 1/an = 0 is what it shoud say say sorry. where an is a sequence

4. Dec 6, 2011

### spamiam

What have you tried? We can't help until we see your attempt. The only hint I will give is to write out the definitions. What does $\lim_{n \to \infty} a_n = \infty$ mean? How about $\lim_{n \to \infty} 1/a_n = 0$?

5. Dec 6, 2011

### lugita15

Why don't you first prove that the limit of 1/n as n goes to infinity is zero, and then consider the limit of 1/a_n as a composition?

6. Dec 7, 2011

### gregy6196

i can prove it graphically but i dont know the deffinitions and this is not for assingnment, once someone shows me how its done then i can start my assignments, i need the basics first

7. Dec 7, 2011

### spamiam

You can find the necessary definitions here: http://en.wikipedia.org/wiki/Limit_of_a_sequence

8. Dec 7, 2011

### gregy6196

thanks. can you please show me how to aproach it

9. Dec 7, 2011

### spamiam

Here's the definition for $\lim_{n \to \infty} a_n = \infty$. For any $M > 0$ there is positive integer N such that for any $n \geq N$, $a_n > M$.

Now you write out the definition for $\lim_{n \to \infty} 1/a_n = 0$ and try to see why $\lim_{n \to \infty} a_n = \infty$ implies $\lim_{n \to \infty} 1/a_n = 0$ from these definitions.

10. Dec 7, 2011

### Sina

For any ε positive you should find an N s.t for all n>N |1/an - 0| < ε. You must show that it would suffice to do this for all ε = 1/m where m is an integer (exercise). Then since your sequence goes to infinity, you can find an N s.t for all n>N an > m hence 1/an < 1/m and then you are done (fill in the gaps if you want to understand the definition of limit).

11. Dec 20, 2011

### gregy6196

thanks for this, can you please give a counter example to show the converse is false?

12. Dec 21, 2011

### evagelos

I think the converse is also true,here is a proof:

Let M>0,then (1/M)>0

Since $$lim_{n\to\infty}\frac{1}{a_{n}}=0$$ for all ε>0 there exists a natural No k such that:

for all ,n $$n\geq k\Longrightarrow \frac{1}{a_{n}}<\epsilon$$.

Put $$\epsilon = \frac{1}{M}$$ and we have that :

for all ,n $$n\geq k\Longrightarrow \frac{1}{a_{n}}<\frac{1}{M}\Longleftrightarrow a_{n}>M$$

Hence $$lim_{n\to\infty} a_{n} = \infty$$

13. Dec 21, 2011

### micromass

This step is false:

$$\frac{1}{-2}<\frac{1}{2}$$

but not

$$-2>2$$

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