Definition of Convergence: Can n -> -infinity

1. Aug 21, 2014

Caiti

1. The problem statement, all variables and given/known data
I've been given a question that makes use of 5^(n)*sin(pi*n!)
The question merely asks if the sequence converges, and if so, to determine its limit. Am I right in assuming that this does converge, under the definition, but does so as n-> - infinity?

So basically, I believe this diverges for n-> infinity. However, for n-> - infinity I believe it will converge to 0. Using the definition of convergence, am I allowed to stay it converges?

2. Relevant equations
The definition of convergence:
A sequence converges as n-> infinity if there is an n>N given any E>0 where |U_n - L| < E
or
U_n -> L as n-> infinity

3. The attempt at a solution

The above definition actually changes my mind. It seems to only states that a sequence is converging if it meets conditions and n-> infinity, and so n-> - infinity wouldn't fit the definition. The question definitely doesn't specify what n is tending to, so my question is literally just whether this is classed as convergence or not?

Thank you very much

2. Aug 21, 2014

Simon Bridge

Usually your series starts at n=1 or n=0 and n increments.
Hence, by definition, the series is not defined for n<0.

It is possible to define a series which extends backwards as well as forwards ... so check the definition used for the example.

When you want to see if a series converges for "$n\to -\infty$" as you put it, you are actually saying you want to see what happens when some part of the function describing the series becomes a very large negative number.
i.e. in your example you could write: $y_n = 5^{-n}\sin(\pi (-n)!)$ and then apply the definition.

Of course you can see this is a different series from the other one.

But what is n! when n<0?

3. Aug 21, 2014

Fredrik

Staff Emeritus
This statement is not correct. Where you say "if there's an n>N", you should be saying "if there's a positive integer N such that for all n≥N, ".

There's a version of the definition that's both easier to understand and easier to remember. Let S be a sequence. A number x is said to be a limit of S, if every open interval with x at the center contains all but a finite number of terms of S.

The ε-N statement of the definition of "limit" is only meant to remove any possible misunderstandings about what "all but a finite" means: A number x is said to be a limit of the sequence $(x_n)_{n=1}^\infty$, if for all ε>0, there's a positive integer N such that for all positive integers n,
$$n\geq N\ \Rightarrow\ |x_n-x|<\varepsilon.$$
Note that to say that $|x_n-x|<\varepsilon$, is equivalent to saying that $x_n$ is in the open interval $(x-\varepsilon,x+\varepsilon)$.

A sequence is said to be convergent if it has a limit.

Is there an open interval that contains all but a finite number of terms of this sequence?

4. Aug 21, 2014

Fredrik

Staff Emeritus
A sequence either converges or it doesn't. So it doesn't make much sense to say that it converges as $n\to\infty$. It's especially weird when you haven't said explicitly what n is. The statement
$x_n\to x$ as $n\to\infty$​
and in particular the notation used in it, is meant to say both that x is a limit of the sequence $(x_n)_{n=1}^\infty$, and that the value of the $n$th term of this sequence (which is denoted by $x_n$) approaches the value of x, as you keep increasing the value of n.

5. Aug 21, 2014

Ray Vickson

It certainly makes sense to say that a two-sided sequence $(x_n)_{n=-\infty}^{\infty}$ converges at one end but not at the other. However, the lack of a definition of n! for n < 0 means that the current sequence does not fall into this category.

6. Aug 21, 2014

Fredrik

Staff Emeritus
Of course, but most books wouldn't even define a "two-sided sequence". They usually just define a sequence as a function whose domain is $\mathbb Z^+$ (positive integers) or $\mathbb N$ (natural numbers including 0).

7. Aug 21, 2014

Ray Vickson

Well, maybe textbooks do not define them, but I have seen them used occasionally, primarily in connection with probability and stochastic processes.

8. Aug 21, 2014

Simon Bridge

You get it in time-sampling models for eg. ... a signal may have existed for some time before t=0 on the stopwatch. $y_n=y(t_n):t_n=n\Delta t$ you end up with a series indexed by n, where n needs to be able to be a negative number for sample data in the "past".

You can always shift the zero to recover the usual proceedings, it's just often convenient to be able to back-project a model.

Normally we would work out what question we want to answer and then figure out how to ask it in maths.
If, as in this case, you start with the maths, the trick is to figure out what question the maths is trying to ask.