# Convergence of sequence using definition

• Mar 24th 2013, 05:24 AM
learning
Convergence of sequence using definition
"using the definition (epsilon-N definition), show that the sequences (-1)^n(1/3n) + (1/2n) and cos(n^2)/(3n+1) both converge to the limit zero". ok, so i know the limit and i know the process. i can handle these questions ok when it is just a polynomial sequence but, for example, what do i do with the (-1)^n? the first step i need to do is deal with this and i don't know how to proceed. similar for cosines.
• Mar 24th 2013, 05:27 AM
Prove It
Re: Convergence of sequence using definition
Can you show us what you tried please?
• Mar 24th 2013, 06:37 AM
learning
Re: Convergence of sequence using definition
hi,
as i said, i encountered a problem with the first step. if someone can tell me how to deal with the (-1)^n, i can do the rest. but i cant try anything beyond that because the obstacle i need help with is right at the beginning..
• Mar 24th 2013, 09:40 AM
Plato
Re: Convergence of sequence using definition
Quote:

Originally Posted by learning
hi,
as i said, i encountered a problem with the first step. if someone can tell me how to deal with the (-1)^n, i can do the rest. but i cant try anything beyond that because the obstacle i need help with is right at the beginning..

For the first one separate it.
$\left| {{{( - 1)}^n}{{\left( {\frac{1}{3}} \right)}^n} + \frac{1}{{2n}}} \right| \le \left| {{{( - 1)}^n}{{\left( {\frac{1}{3}} \right)}^n}} \right| + \left| {\frac{1}{{2n}}} \right|$

Make each of those terms less than $\frac{\varepsilon }{2}$.
• Mar 24th 2013, 11:20 AM
learning
Re: Convergence of sequence using definition
hmmm its 1/3n, not (1/3)^n. so does this make a difference? my thought was that i should just say that the whole thing is less than (something?) regardless of sign so i.e. take the value for even n and say its less than that?
• Mar 24th 2013, 11:35 AM
Siron
Re: Convergence of sequence using definition
To prove $\lim_{n \to \infty} (-1)^n \frac{1}{3n}+\frac{1}{2n}=0$, i.e $\forall \epsilon>0, \exists P>0: |n|>P \Rightarrow \left|(-1)^n \frac{1}{3n}+\frac{1}{2n}\right|<\epsilon$
Proof:
Let $\epsilon>0$ then we have
$\left|(-1)^n \frac{1}{3n}+\frac{1}{2n}\right| = \left|\frac{1}{3n}\right|+\left|\frac{1}{2n}\right |=\left|\frac{1}{n}\right|\frac{5}{6}$.
Since $|n|>P \Rightarrow \left|\frac{1}{n}\right|<\frac{1}{P}$.

Can you finish the proof?
• Mar 24th 2013, 01:50 PM
learning
Re: Convergence of sequence using definition
what i did is:

$|(-1)^n\frac{1}{3n}+\frac{1}{2n}|=|\frac{(-1)^n2n + 3n}{6n^2}| \leq |\frac{5n}{6n^2}| \leq |\frac{5}{6n}|$ etc.... is this the same/an ok way of expressing it? basically saying that 'it has to be less than what it is if n is even', as that gives the highest absolute value. i mean its technically true, do i have to be more rigid in the conditions here? it leads to same the resut. and yes i am confortable with the rest of the proof it was just how to deal with the (-1)^n i have never come across. basically the differnce is i put the two parts over a common denomnator instead of splitting. in fact once the (-1)^n is gone i can just get rid of the modulus altogether since its all always positive then.
• Mar 24th 2013, 02:24 PM
Siron
Re: Convergence of sequence using definition
Can you give the proof for the other limit now?