Jimmy R.

You are welcome! I enjoyed the communication too!

Yes right. I fell disappointed. I think that I am wasting your time, so I will delete my answer. I cannot think of something different that works.

Since $\lim_{n\to \infty}h_n=0$ we know $\limsup_{n\to\infty} h_n=\liminf_{n\to\infty} h_n=0$. But is it perhaps that $\lim_{n\to\infty}\sup_{1\le k\le n}h_n$ is a different thing than $\limsup_{n\to\infty}h_n$? Sorry, for asking all the time, this confuses me.

The last chance is monotone convergence theorem: Let $g_n:=sup_{1\le k\le n}h_n$. Then the $g_n$ are non-decreasing and $\lim g_n=0:=g$. Hence $\lim \int g_n =\int \lim g_n=0$. What do you think about that?

But h_n go to 0. USing this there must be some way to bound them.

@user3825755 Yes right. I do not see how can this work now.

@user3825755 Yes, right. What about the bounded convergence theorem as I said above? It seems to me that this works

@user3825755 Wait, I was certain that it was ok at the moment, but know that you say it, I get some doubts. When I think again over it, it seems that bounded convergence theorem works instead: The $h_n$ are uniformly bounded by $g=1$ and they converge pointwise to $0$, i.e. $h_n(t)\to 0$ for all $t$ as $n\to \infty$ ($t$ plays no role, they are constant). Does this make more sense? (p.s. I was sure, Reverse Fatou worked, but either now or back then I missed something).

@user3825755 It would suffice to define such a $g$, but why is $h_n\le g$?

@Math1000 Good question, I do not know what the implications are? Not to mention +1 for your neat effort (from which my answer directly follows as I mentioned).

It is exactly what you are looking for

Did you check Spitzer's formula?

But perhaps someone else does.

(somehow I delete my Hi!! :) comment). Yes, I do not know of a simple expression

I think you should search in queuing theory books as Did suggested

Χμμμ, έχω κάνει άλλα δεν το χρησιμοποιώ συχνά. Έχεις να γράψεις πρόγραμμα?

δε νομίζω ότι γίνεται αλλιώς, αλλά δεν είμαι και ειδικός στους αλγόριθμους...

Έλα, δες εδώ: en.wikipedia.org/wiki/Greedy_algorithm Στη φωτό δεξιά έχει ακριβώς αυτό το παράδειγμα. Σωστή είσαι

(νομίζω πώς είναι όντως άπληστος, πάντως)

Άλλη ερώτηση: είναι σίγουρα άπληστος αυτός ο αλγόριθμος?

Δηλαδή ακριβώς αυτό που κάνει ο αλγόριθμός σου

Αν θέλεις να κάνεις συναλλαγή για ν σέντς τότε είναι προφανές (ή το δείχνεις εύκολα) ότι ο ελάχιστος αριθμός κερμάτων που χρειάζονται προκύπτει με το να διαιρέσεις τον αριθμό αυτό με το μεγαλύτερο σε αξία νομίσμα που χωράει (π.χ. μιά συναλλαγή 7 σέντς χωράει 1 νόμισμα των 5 σέντς) και μετά με το υπόλοιπο το ίδιο κ.ο.κ.

Για να δείξεις ότι η λύση είναι βέλτιστη μπορείς να βρεις τη βέλτιστη λύση και να διαπιστώσεις ότι είναι ίδια με αυτή που δίνει ο αλγόριθμος

Σωστό μου φαίνεται.

πες μου, ναι, αλλά δεν ξέρω αν μπορώ να σε βοηθήσω.

ναι είχα, αλλά όχι πολλά! τι ακριβώς θες?

Να μιλάμε ελληνικά?

In statistics. At which university are you an undergraduate?

Hmmm, yes and no. I am a PhD student but I am not in the typical age of a student. I am 33 now. Do you just live in greece or are you greek? Are you a student "too" :)?

Athens! Have you been in Greece? Where are you from?

@KhallilBenyattou Unfortunately \cancelout does not work either. Thanks for the suggestion. I am also quite sure that I have seen something like \cancel or \cancelto and I do not know why it does not work now. Perhaps there something wrong with my computer?

@N3buchadnezzar But, what if the oldest has an answer that does not appear in the newest? Actually it would be nice to have an option to merge questions.

That is a guess, I did not check the answers

@N3buchadnezzar Perhaps the reason is that both got nice and different answers. Now which one should one delete?

Hi, I am trying to use the command \cancel in my answers, but it does not work. \not gives a very short line. But I am sure there are some answers around where you can see terms being nicely cancelled out, I only cannot find them now to copy the code. Does anyone know a command that works. I would be grateful because I need it quite often.

@Lord_Farin No problem, I promoted it.

Ok, thanks @Najib @Jimmy. So now we have one possible for each candidate!

Hi everyone! So, maybe silly question - I just landed from Mars: How many votes can we cast in the primary phase that is ongoing now?

(but not unique degenerate)

This is exactly it. Degenerate in primal but unique solution in the dual.

Yes, nice link. This is exactly a counterexample to your professor's question. So you need also uniqueness no matter what.

In the first case, I copied the proof and in the second case you cannot have multiple solutions in the dual, so in this case your question makes no sense. Your professor means - as the homework you quoted - if you have a degenerate solution in the primal then .. bla bla, where this a stands for unique. If not, then his question is wrong.

Your question is answered because: Case 1: primal has unique and degenerate and Case 2: primal has multiple and degenerate. Both cases are treated in my answer.

I am giving it to that question, I know that other users might get annoyed but know we decided it.

U there?

From the ones you posted, only the one about Bolzano-Weierstrass is ok, but you say by yourself that is merely a guess. So, are you ok with the one I suggested?

I think of this one, math.stackexchange.com/questions/957701/…

U there?

Suggest an answer of yours, which you think is best.

Hahaha, ok. I do not really get your logic but it is fine with me ;)