liminf and limsup of a sequence

Question

I came across the following problem:

Find $\text{lim inf} \ a_n$ and $\text{lim sup} \ a_n$ for the following sequence: $a_n = (-1)^{n}+1/n$.

So define $A_n = \{a_{n+1}, a_{n+2}, \dots \} = \{a_k: k \geq n \}$. I think $$A_{n} = \begin{cases} 1+ \frac{1}{n} \ \ \ \ \ \text{if} \ n \ \text{is even} \\ -1 + \frac{1}{n} \ \ \text{if} \ n \ \text{is odd} \end{cases}$$

Now define $$\text{lim sup} \ a_n = b = \inf \limits_{n \geq 1} \ b_n = \inf_{n \geq 1} \left(\sup\limits_{k \geq n} \ a_k \right)$$ and $$\text{lim inf} \ a_n = c = \sup\limits_{n \geq 1} \ c_n = \sup_{n \geq 1} \left(\inf\limits_{k \geq n} \ a_k \right)$$

where $b_n = \sup A_n$ and $c_n = \inf A_n$. I think $b_n = 1+\frac{1}{n}$ for even $n$ and $c_n = -1+\frac{1}{n}$ for odd $n$. Thus $$\text{lim sup} \ a_n = \inf\limits_{n \geq 1} \ \left(1+ \frac{1}{n} \right)= 1$$ and $$\text{lim inf} \ a_n = \sup\limits_{n \geq 1} \ \left(-1+ \frac{1}{n} \right)= 0$$

It follows that the sequence is not convergent. Would this be correct?

Edit. For $n$ even, $a_2 = 1+ \frac{1}{2}$, $a_4 = 1+ \frac{1}{4}$ etc...For $n$ odd, $a_1= 0$, $a_3= -1+ \frac{1}{3}$, etc...

So I get the correct infimum but the wrong supremum. Do I drop $a_1$, $a_3$, etc...? For example, $a_{240} = 1+ \frac{1}{240}$ and $a_{241} = -1+\frac{1}{241}$. So I only look at $A_n$ for large $n$?

Answer 1

You ask whether or not your reasoning is correct. It is close, but not quite there.

Let's start from the beginning. We have $a_n = (-1)^n + 1/n$. Now, we can define (as you did), the sequence of sequences $$A_n = \{a_k\colon k \geq n\} = \{a_n, a_{n+1}, \ldots\}.$$ Note that each $A_n$ is not a number, but rather a sequence of numbers. (This makes your guess for $A_n$ incorrect.)

Now, by definition $$b = \limsup a_n = \inf_{n\geq 1}\ \sup_{k\geq n} a_k = \inf_{n\geq 1}\sup\{a_k\colon k \geq n\} = \inf_{n\geq 1}(\sup A_n) = \inf_{n\geq 1}\, b_n$$ $$c = \liminf a_n = \sup_{n\geq 1}\ \inf_{k\geq n} a_k = \sup_{n\geq 1}\inf\{a_k\colon k \geq n\} = \sup_{n\geq 1}(\inf A_n) = \sup_{n\geq 1} \,c_n,$$ where we let $b_n = \sup A_n$ and $c_n = \inf A_n$, as you did.

So let's compute $b_n = \sup A_n$ and $c_n = \inf A_n$.

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Claim: We claim that $$b_n = \sup A_n = \begin{cases} 1 + \frac{1}{n} \ \ \ \text{ if } n \text{ is even} \\ 1 + \frac{1}{n+1} \ \ \ \text{ if } n \text{ is odd} \end{cases}$$ and $$c_n = \inf A_n = -1 \ \ \text{ for all } n \geq 1.$$

Proof: Let's look at $b_n$ first. If $n$ is even, then $$A_n = \left\{1 + \frac{1}{n}, \ -1 + \frac{1}{n+1}, \ 1 +\frac{1}{n+2}, \ldots\right\}.$$ It is clear that $A_n$ contains a largest element, namely $1 + \frac{1}{n}$. (If you don't find this fully precise, I invite you to fill in the details.) Therefore, $\sup A_n = 1 + \frac{1}{n}$ when $n$ is even. If $n$ is odd, then $$A_n = \left\{-1 + \frac{1}{n}, \ 1 +\frac{1}{n+1}, \ -1 + \frac{1}{n+2}, \ldots\right\}.$$ Again, $A_n$ contains a largest element, namely $1 + \frac{1}{n+1}$, so that $\sup A_n = 1 + \frac{1}{n+1}$. This verifies the claim about $b_n$.

Let's look at $c_n$ now. We want to show that $\inf A_n = -1$, i.e. that $-1$ is the greatest lower bound of $A_n = \{a_k\colon k\geq n\}$.

Now, since $-1 < a_k$ for every $k \geq n$ (you can check this), it follows that $-1$ is a lower bound of $A_n$. To show that it is the greatest lower bound, we need to show that if $q > -1$, then $q$ is not a lower bound for $A_n$.

Let $q > -1$. Choose $k \geq n$ large enough so that $k > \frac{1}{q+1}$. Then $k(q+1) > 1$, so $q+1 > \frac{1}{k}$, so $q > -1 + \frac{1}{k}$. But $-1 + \frac{1}{k}$ is an element of $A_n$ (since $k \geq n$), and we just showed that it is less than $q$. Therefore, $q$ is not a lower bound for $A_n$. We therefore conclude that $\inf A_n = -1$ (for all $n \geq 1$).

This proves the claim.

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Finally, it follows from our claim (I leave the details to you) that $$b = \limsup a_n = \inf_{n\geq 1}\ b_n = 1$$ and $$c = \liminf a_n = \sup_{n\geq 1} \ c_n = -1.$$

Answer 2

Note that for any set of real numbers $S\subset \mathbb{R}$, if $S=X\cup Y$, then $$\inf(S)=\min\{\inf(X),\inf(Y)\},$$ because $a\leq s$ for all $s\in S$ if and only if $a\leq x$ for all $x\in X$, and $a\leq y$ for all $y\in Y$, so $$\{a\in\mathbb{R}\mid a\leq \inf(S)\}=\{a\in\mathbb{R}\mid \forall s\in S, a\leq s\}=$$ $$\{a\in\mathbb{R}\mid \forall x\in X,a\leq x\}\cap\{a\in\mathbb{R}\mid \forall y\in Y,a\leq y\}=$$ $$\{a\in\mathbb{R}\mid a\leq\inf(X)\}\cap\{a\in\mathbb{R}\mid a\leq\inf(Y)\}=\{a\in\mathbb{R}\mid a\leq\min\{\inf(X),\inf(Y)\}\}.$$

A similar argument shows that $$\sup(S)=\max\{\sup(X),\sup(Y)\}.$$

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Claim: $\liminf a_n=-1$.

Note that for any even $p$ and odd $q$, $$a_p=1+\frac{1}{p}\geq -1+\frac{1}{q}=a_q.$$ Thus, if $X\subset\mathbb{N}$ is a set containing only even numbers and $Y\subset\mathbb{N}$ is a set containing only odd numbers, then $$\inf_{m\in X}\;a_m\geq\sup_{m\in Y}\;a_m.$$ Certainly, this implies that $$\inf_{m\in X}\;a_m\geq\inf_{m\in Y}\;a_m.$$

By definition, $$\liminf_{n\to\infty}\;a_n = \lim_{n\to\infty}\Big(\inf_{m\geq n}a_m\Big).$$

By our result above, we have that $$\liminf_{n\to\infty}\;a_n = \lim_{n\to\infty}\Big(\min\Big\{\inf_{\text{even } m\geq n}a_m,\;\;\;\; \inf_{\text{odd } m\geq n}a_m\Big\}\Big).$$

But $$\min\Big\{\inf_{\text{even } m\geq n}a_m,\;\;\;\; \inf_{\text{odd } m\geq n}a_m\Big\}=\inf_{\text{odd } m\geq n}a_m,$$ so $$\liminf_{n\to\infty}\;a_n=\lim_{n\rightarrow\infty}\Big(\inf_{\text{odd } m\geq n}a_m\Big)$$ Now prove that $$\inf_{\text{odd } m\geq n}a_m=\inf\{-1+\tfrac{1}{m}\mid \text{odd }m\geq n\}=-1,$$ so that $$\liminf_{n\to\infty}\;a_n=\lim_{n\rightarrow\infty}(-1)=-1.$$

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A dual argument of everything above will show that $\limsup a_n=1$.

Answer 3

You are right in that if $\limsup a_n$ and $\liminf a_n$ are different from each other, then the sequence, as a sequence of real numbers, cannot be convergent, since both $\limsup$ and $\liminf$ are limit points of the sequence, and, by the triangle inequality, a sequence cannot have more than one limit (the uniqueness of the limit of a sequence holds, more generally,for metric spaces,and, even more generally, to Hausdorff spaces).

To throw some more in, $\limsup$ and $\liminf$ are guaranteed to exist in the extended real numbers, since they are the limits of monotone sequences given by the collection of sups and infs; taking sups over increasingly-larger sets gives you a non-decreasing sequence, and similar for the sequence of infs; if these sequences are bounded, then they will converge to a real number, and, if the sequences associated to each are unbounded, then they will converge to $\pm\infty$ respectively.

You can also see $\limsup$ and $\liminf$ as the largest and smallest limit points of the sequence. Finally, to get to your case, I think Wikipedia has a nice rule used to determine what your $\limsup$ and $\liminf$ are: $\limsup a_n$ is the largest value $b$ (to completely ripoff Wikipedia) of $a_n$ such that for each $\epsilon\gt 0$, only finitely many terms of the sequence are larger than $b+\epsilon$. Check to see if: 1) $1$ is a limit point of $\{a_n\}$, i.e., does every neighborhood of $1$ contain infinitely many points of the sequence; and 2) Is it the case that for any $\epsilon>0$ there are only finitely many terms larger than $1+\epsilon$. Do something similar for $a=-1$; check that it is a limit point of $\{a_n\}$, and check also that for every $\epsilon>0$, there are only finitely-many points smaller than $a-\epsilon$.

So, for $1$, for any $\epsilon>0$, there is , by the Archimedean Principle, an integer $N$ with $\frac {1}{N}\lt\epsilon$.

So all terms $1+\frac {1}{n}$ with $n\gt N$ will be smaller than $1+\epsilon$. Then every neighborhood of $1$ contains infinitely many points of $\{a_n\}$, and, by the above paragraph, $1$ is the $\limsup$ of $\{a_n\}$.

Similarly , for $-1$, using the Archimedean principle again, there will be an $N'$ such that $\frac {1}{n} \lt\epsilon$ for all $n>N'$ and so all terms $-1+\frac {1}{n}$, with $n>N'$ will be larger than $-1+\epsilon$.

But notice that your choice of $0$ will not do. Take a neighborhood of $0$ with $\epsilon=\frac {1}{2} $. Then, since both sequences are monotone (because $n>n'$ impies $\frac {1}{n} <\frac {1}{n}$ ) and $a_2= \frac{3}{2}$, while $a_3=\frac{-2}{3}$, there will be no terms at all of the sequence in the neighborhood $(\frac{-1}{2},\frac{1}{2})$ of $0$. So $0$ is not a limit point of $a_n$, let alone $\liminf a_n$.

So, when it comes to $0$: no Sup for you!

Answer 4

Here is a useful result. Its proof is a nice problem

If $\{a_n\}$ is a sequence then $$\liminf_{n\to\infty} a_n$$ is the smallest limit point of the sequence and $$\limsup_{n\to\infty} a_n$$ is the largest. Apply this result for a quick solution to your problem.