Proving the Limit of a Sequence Using Formal Definition

[Telemachus]

Hi there, I'm new around here. I got this problem, I must demonstrate next limit by using the limits formal definition.
a_n= \displaystyle\frac{(-1)^n}{n^2}, L=2

Which means:

\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{(-1)^n}{n^2}}=2 (its a sequence, n\in{\mathbb{N}}).

And I don't know how to solve this. Here is the formal definition of limit given on class:

\displaystyle\lim_{n \to{+}\infty}{a_n}=L\Leftrightarrow{\forall\epsilon>0\exists{n_0=n_0(\epsilon) \textsf{ such that } n\geq{n_0}\Rightarrow{|a_n-L|<\epsilon}}}I've tried to solve it, but couldn't find the way. I don't know how to solve (-1)^n. I actually thought at first sight that it didn't has a limit because of it, cause (-1)^n oscillates, but I actually don't know.

Thank you for reading. Bye bye.

 

[ystael]

Well, the result you are trying to prove is false; that may be part of the difficulty. Did you read the question correctly?

When you have a sequence whose behavior you aren't sure of, it usually helps to write out the first few terms (three to ten, depending on how hard they are to compute) and see if you can see any patterns. In this case that should give you an idea what the limit is and what the sequence's behavior is like.

 

[Telemachus]

Well, the result you are trying to prove is false; that may be part of the difficulty. Did you read the question correctly?

When you have a sequence whose behavior you aren't sure of, it usually helps to write out the first few terms (three to ten, depending on how hard they are to compute) and see if you can see any patterns. In this case that should give you an idea what the limit is and what the sequence's behavior is like.

Thank you for answering ystael. I supposed it was false, I must now prove that it is false I think. I'll still working on it. Thank you again.

 

[Telemachus]

Well, hi there again. I get to this result, but I don't know if it's a complete demonstration, or if its well done.

\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{(-1)^n}{n^2}}=2\Leftrightarrow{\forall\epsilon>0\exists{n_0=n_0(\epsilon) \textsf{ such that } n\geq{n_0}\Rightarrow{|\displaystyle\frac{(-1)^n}{n^2}-2|<\epsilon}}}

\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{(-1)^n}{n^2}}=2\Leftrightarrow{\forall\epsilon>0\exists{n_0=n_0(\epsilon) \textsf{ such that } n\geq{n_0}\Rightarrow{|\displaystyle\frac{(-1)^n}{n^2}-2|<\epsilon}}}\not{\exists}n_0(\epsilon)\textsf{ such that }n\geq{n_0}\Rightarrow{|\displaystyle\frac{(-1)^n}{n^2}-2|<\epsilon}\therefore\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{(-1)^n}{n^2}}\neq{2}

Is this ok as a demonstration that 2 is not a limit of a_n?

 

[Dick]

Well, hi there again. I get to this result, but I don't know if it's a complete demonstration, or if its well done.

\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{(-1)^n}{n^2}}=2\Leftrightarrow{\forall\epsilon>0\exists{n_0=n_0(\epsilon) \textsf{ such that } n\geq{n_0}\Rightarrow{|\displaystyle\frac{(-1)^n}{n^2}-2|<\epsilon}}}

\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{(-1)^n}{n^2}}=2\Leftrightarrow{\forall\epsilon>0\exists{n_0=n_0(\epsilon) \textsf{ such that } n\geq{n_0}\Rightarrow{|\displaystyle\frac{(-1)^n}{n^2}-2|<\epsilon}}}\not{\exists}n_0(\epsilon)\textsf{ such that }n\geq{n_0}\Rightarrow{|\displaystyle\frac{(-1)^n}{n^2}-2|<\epsilon}\therefore\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{(-1)^n}{n^2}}\neq{2}

Is this ok as a demonstration that 2 is not a limit of a_n?

You didn't really do anything except say that there doesn't exist an n0 without proving it doesn't exist. Try thinking about this more concretely. Can you find an n0 such that for all n>n0, |an-2|>1? That would prove it, right?