Lim sup and lim inf the averages of a sequence

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In summary: M*(s1+s2+...sN) + lim sup sn <= lim sup sn. This proves the second part of the inequality.In summary, we have shown that lim inf sn <= lim inf \sigman and lim sup\sigman <= lim sup sn. This proves the given inequality. I hope this helps you understand the problem better. Let me know if you have any further questions.
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Homework Statement


let sn be a non-negative sequence, and define [tex]\sigma[/tex]n to be 1/n*(s1+s2+...sn)

Show that lim inf sn <= lim inf [tex]\sigma[/tex]n and lim sup[tex]\sigma[/tex]n <= lim sup sn

Homework Equations



Hint: For the last inequality show that M>N imply: sup{[tex]\sigma[/tex]n:n>M} <= 1/M*(s1+s2+...sN) + sup{sn:n>N}


The Attempt at a Solution



So one fact I could gather is that [tex]\sigma[/tex]n <= Max{s1,s2...sn} since the average can't be higher than the highest member value. Also for M>N, let sup{s_n:n>N} = sm, then the average is maximized at m, and if m<M then sup{[tex]\sigma[/tex]n:n>M} < [tex]\sigma[/tex]m, since all elements of the set of averages are forced to take elements smaller than sm such as sm+1, sm+2... sM. But am I on the right track? And can I use this information to deduce what was given in the Hint.

Also, even when I have the hint how would I connect it with lim sup? In other words, what would be the index going to infinity for the sequence 1/M*(s1+s2+...sN)?
 
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Thank you for your post. After reviewing the problem, I would like to provide you with a solution to help you better understand the concept.

First, let's define the terms lim inf and lim sup. The lim inf of a sequence is the greatest lower bound of the set of all subsequential limits, while the lim sup is the least upper bound of the set of all subsequential limits. In simpler terms, the lim inf is the smallest value that the sequence can get infinitely close to, while the lim sup is the largest value that the sequence can get infinitely close to.

Now, let's prove the given inequality. We will start with the first part: lim inf sn <= lim inf \sigman. To prove this, we will use the definition of lim inf and lim sup. Since sn is a non-negative sequence, we can say that lim inf sn is the smallest value that sn can get infinitely close to. Similarly, 1/n*(s1+s2+...sn) is the average of the first n terms of the sequence, so lim inf \sigman is the smallest value that this average can get infinitely close to. Since the average can't be higher than the highest member value, we can say that lim inf \sigman <= Max{s1,s2...sn}. This implies that lim inf \sigman <= lim inf sn, which proves the first part of the inequality.

Next, let's prove the second part: lim sup\sigman <= lim sup sn. To prove this, we will use the hint given in the problem. Let M>N. This means that the supremum of the set of averages {sigman:n>M} is less than or equal to the average of the first N terms of the sequence. In other words, sup{\sigman:n>M} <= 1/M*(s1+s2+...sN). We can also say that the supremum of the set of all subsequential limits of sn is equal to lim sup sn. This means that sup{sn:n>N} = lim sup sn. Combining these two facts, we get sup{\sigman:n>M} <= 1/M*(s1+s2+...sN) + sup{sn:n>N}. Since this holds for all M>N, we can say that lim sup\sigman <= 1/M*(s1+s2+...sN) + lim sup sn. As M approaches infinity
 

1. What is the definition of "lim sup" and "lim inf"?

The lim sup (limit superior) of a sequence is the largest value that the sequence approaches as n approaches infinity. The lim inf (limit inferior) of a sequence is the smallest value that the sequence approaches as n approaches infinity.

2. How do you calculate the lim sup and lim inf of a sequence?

To calculate the lim sup and lim inf of a sequence, you can take the supremum (smallest upper bound) and infimum (largest lower bound) of the set of all possible limit points of the sequence, respectively. Alternatively, you can also use the definitions of lim sup and lim inf and check for convergence and divergence of the sequence.

3. What is the significance of lim sup and lim inf in a sequence?

The lim sup and lim inf of a sequence give important information about the behavior of the sequence as n approaches infinity. They can help determine whether a sequence converges or diverges, and if it converges, what value it converges to.

4. Can the lim sup and lim inf of a sequence be equal?

Yes, the lim sup and lim inf of a sequence can be equal if the sequence is convergent. In this case, both the lim sup and lim inf will have the same value, which is the limit of the sequence.

5. How do lim sup and lim inf relate to each other in a sequence?

It can be shown that the lim inf is always less than or equal to the lim sup in a sequence. Furthermore, if the lim sup and lim inf are equal, then the sequence is convergent. If they are not equal, then the sequence is divergent.

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