@Hawk No. Dirichlet's test is about series of the form $\sum a_n b_n$. We need to look at the sequence $(a_n)$ and the partial sums $\sum_{n=1}^k b_n$ separately to verify that the premises of the test are given, but we don't treat $\sum a_n$ and $\sum b_n$ separately, in general, neither converges.
@PedroTamaroff Yes it is. That's why I dropped in - and I'm guesing it's why Jack dropped in too - to tidy up and remove the noise, and curb its source.
@Hawk I intended the question "Who's spamming?" to be taken as something of a warning. I also 'kicked' the user to give them a chance to voluntarily cool off, but I didn't feel I got a helpful response. Does that make sense?
@JackDouglas Definitely that makes sense, sorry, I did not mean to offend you, but I was not aware of your line of action....but I should have trusted that a moderator will take actions much sensibly...thank you for responding.
So it suffices you find the limit $a_{n+1}/a_n$ of $a_n=n^n/n!$, since you have $\log a_n^{1/n}$. But that is easy, since the quotient is $(n+1)^n/n^n$.
I wonder if there is a very fast way of testing the convergence of $$\sum_{n=1}^{\infty} \frac{\displaystyle \sin\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)}{n\displaystyle \left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)}$$
This would give $\log n!\ge (n-1)\log n$, and then $n!\ge \frac{n^n}{e^n}$. Which does not seem strong enough for the mirgee's problem. So this is probably not what Ted had in mind.
@Gabriel: Glad to hear it. I'm hoping to get the first half of the course videoed next fall, my last time teaching it. It was really intended more for my own students, but I'm glad it's of interest.
@Martin: What I intended was $\log n! = \sum_{k=1}^n \log k > \int_1^n \log x\,dx$.
@Pedro: Laziness does not you become :P Tennis time!
If it's the sequence, I think Stirling might be needed. But my approach shows, I believe, that $$\frac1e < \left(\frac ne\right)^n\cdot\frac1{n!} < e.$$
@Hawk Not as free as we were in the seventies and eighties, but still relatively free, comparing to the rest of the world. Or did you mean whether I have time at the moment?
@Hawk The $$\sum_{n=1}^\infty \frac{\sin H_n}{nH_n}$$ one? Don't know, would have to think about it. See whether $H_n = \log n + \gamma + O(n^{-1})$ leads somewhere first, I guess.
@DanielFischer I do not understand fully, so I open conditional convergence on wikipedia...and it says, that $\sum\limits_{n=0}^{\infty}|a_n|=\infty$ when there is conditional convergence.
@Hawk If $\sum \lvert a_n\rvert < \infty$, the series would be unconditionally convergent, so if you have a truly conditionally convergent series, it doesn't converge absolutely.
