@TedShifrin The undergrad chair has already reached out to talk to me. I think I just have to wait for it to play out. The consequences will be the consequences.
@TedShifrin Yeah, I probably did. I wasn't really hired by a person, but a committee. I'm going to lose any chance of a full-time job there (I probably have already). I'm in a full-blown panic. Anyway, thanks!
@TedShifrin I started doing that already. But she's lied about me. And the situation isn't quite what you think. I"m not a grad student, I'm a part time employee with a not-very-strong master's degree.
Yeah. Well, I had sent her an email once about the issue, including taking some of the blame (I'm not blameless), but she never responded. Today we had emails with open hostility, and she copied the coordinator, blaming me for mistakes and bad instructions that she gave. I responded in kind, and then she elevated to the department char. (Though I'm sure you don't care about that)
(Unless you finally made it to work) How can $n$ vary continuously? Also, what is the comparison using u-sub? From this exercise I get that $$\int_1^2 \frac 1x dx \leq a_n \leq \int_1^2 \frac 1x dx$$ and from the other exercise (over the interval from [n, 2n]) I found that $$\int_{n+1}^{2n+1} dx \leq a_n \leq \int_{n}^{2n} dx.$$ All the integrals resolve easily to $\ln(2)$.
Mike gave it to me. I was sort of thinking along those lines when he posted it (I almost wish he'd waited about a few minutes -- but then I also need to get this done (which does not explain why I'm busy making a pretty picture in Desmos))
@MikeMiller I'm making a picture (because I'm in capable of visualizing anything until I draw it AND program a picture). But it's pretty clearly going to work out nice. I thought you were going to work, but if you're utterly bored, just change the value of $n$ https://www.desmos.com/calculator/x56jpttjbn
@MikeMiller OK, when n=2, then k= 3, 4 and the first box's feet are 3/2, 4/2. The second box' feet are 4/2, 5/2. I get the idea, but those feet aren't within [1,2], so maybe we need a sligtly different formula.
@MikeMiller Responding to just this now (then going to look at your other message): There are two sections of the text describing partitioning, but there is no formal definition of right-hand approximation. It's also not in the index.
My boss insists this is not a typo: For the series $\sum_{n=1}^{\infty} a_n$ has terms $a_n = \sum_{k=n+1}^{2n} \frac 1k$, "Show that $a_n$ is equal to the right-hand approximation for the area under $\frac 1x$ over the interval $[1, 2]$". Shouldn't the interval be $[n, 2n]$? How could I compare sum 1/k to interval [1,2]?
So one thing I'm noticing is that $a_n = a_{n-1} - \frac 1n + \frac{1}{2n-1} + \frac{1}{2n} = a_{n-1} + \frac{1}{2n(2n-1)}$ (previous term, less the first fraction, plus the last two). So then $a_n is always increasing.
For the same $$a_n = \sum_{n+1}^{2n} \frac 1k$$ is it sufficient to say that $a_1=1/2$ and $a_n = a_1 + \text{positive terms}$ to prove that $a_n \geq \frac 12$?
