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 $n$?

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    $\begingroup$ You can easily find a subsequence which converges to $-1$, which is less that what you claim is the $\liminf$. So something is amiss... $\endgroup$ – Mariano Suárez-Álvarez Jun 25 '11 at 21:02
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    $\begingroup$ The subsequence $(a_{2n+1})$ tends to $-1$. $\endgroup$ – Américo Tavares Jun 25 '11 at 21:09
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    $\begingroup$ ... and the subsequence $(a_{2n})$ tends to $1$. $\endgroup$ – Américo Tavares Jun 25 '11 at 21:11
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    $\begingroup$ Your definitions need checking - what is $k$ in your definitions of lim sup and lim inf? What you need to be doing is checking the ultimate behaviour of the sequence when you drop off the initial portion, and taking the limit of that. $\endgroup$ – Mark Bennet Jun 25 '11 at 21:34
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    $\begingroup$ Your error in the $\liminf$ is that the infimum of the $a_k$ is not $-1+\frac{1}{n}$. Think about it. What is the infimum of the $a_k$ for $k\geq n$? $\endgroup$ – Arturo Magidin Jun 26 '11 at 3:35

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 $c_n = \inf A_n$.

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.

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 $$c = \liminf a_n = \sup_{n\geq 1} \ c_n = -1.$$


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 $$\sup(S)=\max\{\sup(X),\sup(Y)\}.$$

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 $$\liminf_{n\to\infty}\;a_n=\lim_{n\rightarrow\infty}(-1)=-1.$$

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


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!

  • $\begingroup$ Please give me some time to edit my answer, so I can practice my Latex. $\endgroup$ – gary Jun 25 '11 at 22:38
  • $\begingroup$ I'm done formatting for today; one hour's work is enough. I will redo later, sorry. $\endgroup$ – gary Jun 26 '11 at 0:04
  • $\begingroup$ I hope the long answer is O.K, for me to also practice, even though it may be more than the OP asked for. $\endgroup$ – gary Jun 26 '11 at 0:42
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    $\begingroup$ @gary: (I didn't downvote); are you planning to format this properly, or did you give up in the end? $\endgroup$ – Arturo Magidin Jun 26 '11 at 3:42
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    $\begingroup$ @Arturo: It would be great if you could help. I think I have gone a reasonable way towards learning formatting, but I am kind of tired for now. Feel free, and thanks in advance. $\endgroup$ – gary Jun 26 '11 at 4:38

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.


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