# Basic sequence help. (Convergence)

tamintl
Im struggling with the concept of this basic sequence question.

Let x(n) be a sequence such that lim(n->00) (nx(n)) = 0

i.e. it converges to zero...

How could i show that there is an N s.t. for all n≥N : -1 < nx(n) < 1

Any tips would be great.. I don't want an answer.. I want to be guided through it please.

Regards as ever.
Tam

spamiam
What is the definition (using $\epsilon$ and N) of convergence for a sequence? Once you've written that, it should be straightforward to answer your question.

tamintl
the convergence of a seq:

A sequence {r^n}00n=0 converges if -1< r <1

Same as above but replace 'r^n' with 'nx(n)

Thanks so far

Dr. Seafood
^ Wut?? That's the result of convergence of a geometric series, it's very different.

A sequence of real numbers $(x_n)$ converges when there exists a number $L \in \mathbb{R}$ such that, for any $\epsilon > 0$, I can find a number $N \in \mathbb{N}$ so that $|x_n - L| < \epsilon$ whenever $n > N$. This is very wordy but I'm sure this was introduced in your math class. The number L is the limit of the sequence $(x_n)$.

What do you mean by "nx(n)"? Do you mean $nx_n$?? If so, to show your desired result, use the definition from above. We know that for any number $\epsilon > 0$, I can find $N \in \mathbb{N}$ so that $|nx_n - L| < \epsilon$ whenever n > N. Fill in the details now: we know L = 0. Also, the previous statements work for any chosen $\epsilon$ -- how can you choose $\epsilon$ to get the result?

tamintl
Right:

|x(n) - L| < ε

therefore: |nx(n) - 0| < ε

therefore: nx(n) <

so, -ε < nx(n) < ε

so, take epsilon to be ε=1 and we have: -1 < nx(n) < 1

How does this look guys?

Thanks so far