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.

Re: Convergence of sequence using definition

Can you show us what you tried please?

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..

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.

$\displaystyle \left| {{{( - 1)}^n}{{\left( {\frac{1}{3}} \right)}^n} + \frac{1}{{2n}}} \right| \le \left| {{{( - 1)}^n}{{\left( {\frac{1}{3}} \right)}^n}} \right| + \left| {\frac{1}{{2n}}} \right|$

Make each of those terms less than $\displaystyle \frac{\varepsilon }{2}$.

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?

Re: Convergence of sequence using definition

To prove $\displaystyle \lim_{n \to \infty} (-1)^n \frac{1}{3n}+\frac{1}{2n}=0$, i.e $\displaystyle \forall \epsilon>0, \exists P>0: |n|>P \Rightarrow \left|(-1)^n \frac{1}{3n}+\frac{1}{2n}\right|<\epsilon$

Proof:

Let $\displaystyle \epsilon>0$ then we have

$\displaystyle \left|(-1)^n \frac{1}{3n}+\frac{1}{2n}\right| = \left|\frac{1}{3n}\right|+\left|\frac{1}{2n}\right |=\left|\frac{1}{n}\right|\frac{5}{6}$.

Since $\displaystyle |n|>P \Rightarrow \left|\frac{1}{n}\right|<\frac{1}{P}$.

Can you finish the proof?

Re: Convergence of sequence using definition

what i did is:

$\displaystyle |(-1)^n\frac{1}{3n}+\frac{1}{2n}|=|\frac{(-1)^n2n + 3n}{6n^2}| \leq |\frac{5n}{6n^2}| \leq |\frac{5}{6n}|$ 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.

Re: Convergence of sequence using definition

Can you give the proof for the other limit now?