Proving lim(n→∞)1/an=0 if lim(n→∞)an=0 is False

[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.

 

[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).

 

[gregy6196]

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

 

[spamiam]

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

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?

 

[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?

 

[gregy6196]

i can prove it graphically but i don't 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

 

[spamiam]

i can prove it graphically but i don't 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

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

 

[gregy6196]

thanks. can you please show me how to approach it

 

[spamiam]

thanks. can you please show me how to approach it

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.

 

[Sina]

For any ε positive you should find an N s.t for all n>N |1/a_n - 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 a_n > m hence 1/a_n < 1/m and then you are done (fill in the gaps if you want to understand the definition of limit).

 

[gregy6196]

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

 

[evagelos]

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

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}}&lt;\epsilon.

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

for all ,n n\geq k\Longrightarrow \frac{1}{a_{n}}&lt;\frac{1}{M}\Longleftrightarrow a_{n}&gt;M

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

 

[micromass]

\frac{1}{a_{n}}<\frac{1}{M}\Longleftrightarrow a_{n}>M[/tex]

This step is false:

\frac{1}{-2}&lt;\frac{1}{2}

but not

-2&gt;2