Clement C.

@davidllerenav Yes. We do not want a second derivative to exist, since you already proved that if $f''(0)$ existed, then the sum would converge. So if we hope to find a counterexample, it must be one where $f''(0)$ does not exist.

Just a way to write the term of the "expansion" we get to choose.

@davidllerenav Did that answer your question? If so, consider upvoting the relevant answer(s).

That's why I went for $\phi(x) = \frac{x}{\ln x}$. It satisfies the $x^2 \ll \phi(x) \ll x$ requirement, and (knowing about the classic result $\sum_n \frac{1}{n\ln n} = \infty$) also gives divergence.

@davidllerenav as I mentioned, we wanted to have $f$ such that $f(0)=f'(0)=0$, but $f''(0)$ does not exist. A Taylor expansion around $0$(if $f''(0)$ existed) would be of the form $f(x) = f(0)+f'(0)x+f''(0)x^2 + o(x^2) = f''(0)x^2 + o(x^2)$; since $f''(0)$ doesn't exist, my intuition was that we want something "intermediate", like $$f(x) = f(0)+f'(0)x+\phi(x) + o(\phi(x)) = \phi(x) + o(\phi(x))$$ for $\phi(x) \gg x^2$. We also want the series $\sum_n f(1/n) \approx \sum_n \phi(1/n) $ to diverge, so we cannot have $\phi(x0 = x^c $ for $c\in(1,2)$ (that would give convergence).

@davidllerenav Did that clear your doubts?

@davidllenerenav and log(1/t) = -log(t), so...

@davidllerenav Plug in x=1/n and see...

Because $\sum_n \frac{1}{n \ln n}$ diverges, seither by the Cauchy condensation test, or because of the integral test, since $\int^t \frac{dt}{t\ln t} = \ln\ln x$. @davidllerenav

@davidllerenav It does. $$\frac{x/\ln x - 0}{x-0} \to 0$$ If it didn't, it wouldn't be a counterexample for (2).

A standard diverging series is the one above. I also didn't want a second derivative, since otherwise this falls under a result you already have proven. But a sexond derivative at 0, by a Taylor expansion, corresponds to a $x^2$ term. Something going to 0 "between x and $x^2$" seemed reasonable, hence the log.

Probably. Why do you want another one, though?

The series wouldn't converge. (Note that f(0)=1)

@davidllerenav It's not obvious the series converges with $x\sin\frac{1}{x}$ (it doesn't, I think). Is that what you meant?

@Thomas It's fundamentally different. When the editor makes a call, it's supposedly documented in the system and the editor's responsibility. Asking to change the review makes it look so the reviewers agreed: change not documented, no disagreement on record, editor bearing no responsibility.

@quid I see. Thanks!

(I'm pretty sure serial upvoting is not really in accordance with the rules, but...)

I unexpectedly just got a +100 rep notification. As far as I can tell, it's a 10*(+10), on different answers (not all very recent). Given the sudden thing, it looks like serial upvoting, and I have no idea why -- should I notify you? Is it fishy?

Hi! I have a question

"PS: I'm female" sounds like a terrible idea. First, it may be misinterpreted by the editor or the authors (or is that even remotely clear, if one was oblivious to the use of pronouns?). Second, in many fields where women are underrepresented, it's a potential immediate breach of anonymity...

@shuhalo "Get over it." Whom are you talking to exactly? And regardless of whom: if you are in favor of an "open forum," maybe try not being rude?

@kdog Not that I feel strongly either way, but aren't you misrepresenting what that $13$ is? It's it the difference between up- and down-votes, not the number of upvotes (as far as I know). So it does not reflect "the opinion of the 13 people on the site who upvoted it" -- for all I know (not having access to this statistic yet), there may have been 10001 people voting, with 5007 upvotes and 4994 downvotes.

@verifying The proof has been posted 2 days ago. Either you expect it to be verifiable in two days, in which case the only reasonable option is "because it's blatantly false" (and Kaveh's comment fully applies); or you expect it to be complex and plausible enough so that its correctness is not trivial to establish, in which case... don't hold your breath, and don't wait for an answer before months. Either case, the number of views is irrelevant.

Upvoted; the first paragraph by itself warrants it.

I am downvoting this as (in my opinion) it is opinion-based, sanctimonious, and also utterly unhelpful to the OP.

@DMML a new student asked this by their adviser might not feel like they can say no...

I am not sure how to report this, or if I should (is the behavior mentioned below a violation of the rules?). Basically, an user downvoted an answer of mine because they had misunderstood the question. Since they couldn't apparently undo their downvote, to make up for it (I guess?) they have been through many of my older answers to different questions, and upvoted them.

is there any admin or moderator around?

Yup. Most of the MO posts are a variation of "I am not sure I can parse the title, and when by chance I can it gets answered in a fraction it takes me to read the question"...

@Semiclassical That makes two of us :)

(I have considered Mathoverflow, but I am not sure it fits the bill there -- even though is is a research question. Plus, I don't have much reputation there)

Is it just that no one knows, or is it that bounties are inherently useless given the number of questions appearing on Math.SE?

Basically, I have already put two bounties on it to "draw attention" (the second expires tomorrow), but in vain.

Almost, you have two typos in one equation. It's $$\mathbb{E}[X](a-\mathbb{E}[X]) \leq \frac{a^2}{4}$$ not $$\mathbb{E}(a-\mathbb{E}[X]) \leq \frac{a^2}{2}.$$

It's a definition I chose, exactly for the fact that this gives this inequality. Once this is done, it only remains to show that $f(x) \leq \frac{a^2}{4}$ for all $x\in[0,a]$ (using calculus) to get that $f(\mathbb{E}[X])\leq \frac{a^2}{4}$.

This is what you proved in (b): $\text{Var}[X] \leq \mathbb{E}[X](a-\mathbb{E}[X])$. By definition of $f$, $f(x)=x(a-x)$, so $f(\mathbb{E}[X])=\mathbb{E}[X](a-\mathbb{E}[X])$.

You (i) know that $\mathbb{E}[X] \in [0,a]$, (ii) showed that $\operatorname{Var}[X] \leq f(\mathbb{E}[X])$, and (iii) showed that $f(x) \leq f(\frac{a}{2}) = \frac{a^2}{4}$ for all $x\in[0,a]$. Can you put the pieces together?

Based on that, you should be able to prove (c) as well.

The maximum (by looking where the derivative cancels, you show it's a critical point, but it's a maximum indeed) is attained at $x=\frac{a}{2}$, so the maximum itself is $f(\frac{a}{2}) = \frac{a}{2}(a-\frac{a}{2}) = \frac{a^2}{4}$.

No - use calculus to find it.

(Regarding your current proof -- the implication is $$0\leq X\leq a \Rightarrow 0\leq X^2\leq a X\Rightarrow 0\leq \mathbb{E}[X^2]\leq \mathbb{E}[aX]=a\mathbb{E}[X]$$ not $$0\leq \mathbb{E}[X]\leq a \Rightarrow 0\leq \mathbb{E}[X^2]\leq \mathbb{E}[aX]=a\mathbb{E}[X]$$

For the last question: You know that (i) $0\leq \mathbb{E}[X] \leq a$ and (ii) $\operatorname{Var}[X] \leq \mathbb{E}[X](a-\mathbb{E}[X]) = f(\mathbb{E}[X]))$ for the function $f(x)=x(a-x)$. What is the maximum of $f$ on $[0,a]$?

Yes, basically. What you are using is the property $0\leq Y$ a.s. implies $0 \leq \mathbb{E}[Y]$ (that is, monotonicity of expectation).

Since $0\leq X\leq a$, you have $0\leq X^2 = X\cdot X \leq aX$. Can you use that to derive (a)?

\o/ Let's wait and see.

OK -- thanks!

(added)

I'll add the line (sorry, I forgot). Also, should I add a notice that the title is part of the riddle, or is that implicit?

@Randal'Thor puzzling.stackexchange.com/questions/41451/… (thanks, that's a quick answer!)