Hm, admittedly the answer in question is not of any mentionable quality. But it does at least attempt to answer the question (following the discussion with OP in the comments).
I'm not sure if there is a nice way to compute $\lim_{n\to\infty} n\left(\sum_{k={n+1}}^{\infty} \frac{1}{k^2}\right)$. Apparently Riemann sum works but it requires some additional work.
@robjohn: The Mathematica programmers had only finite time to implement heuristics for their TeXForm command. Therefore it is likely that any given expression generated by said command can be improved upon, semantically and/or syntactically; in fact, the gruesome syntax it produces nearly always puts me off using it.
Hm, I'm only now realising that it may be awkward to speak of the "semantics" of a TeX expression when I refer to the generated output :).
Lord Farin, your question reminds me a question that I solved entirely without calculus (some months ago) $\sum_{n=1}^{\infty}\frac{1}{(4n^3)-4n}=\frac{3}{4}\log 2 -\frac{1}{2}$. Maybe I can repet the story with the question above.
@J.M. I'm thinking of creating an account with mathematica...I have the software, but alas, have not taken advantage of even a mere one percent of its power.
@Chris'ssisterandpals You could properly ask "what's wrong with your head?!" about being a mod... it's a lot of janitorial work. At least, at mathematica.SE, the volume is more manageable.
@amWhy S'okay, the nice questions generate a lot of fine answers.
@J.M. Any tips I should know about the ins and outs of the site? Of course, I should read the FAQ...and search. But is it generally "friendly" and patient with new users?
@skullpatrol Has Jasper appeared in chat at all? I saw is avatar lurking in the background late on the 9th, or maybe it was the 10th, AFTER he deleted.
@caveman I have attention deficit hyperactivity disorders i have to do at least 3 or 4 things the same time, which means actually i don't do anything right at all, which makes me kind of slow in everything :(
@J.M. I think they can be misguided, absolutely. And some believe it's overdiagnosed, not taking into account the impact that our technologically complicated world has had on humans.
Not that there aren't some who REALLY do need help, for which the diagnosis is appropriate...but some find it easier, with children, to deal with restless and bored children by medicating them...
@amWhy "some find it easier, with children, to deal with restless and bored children by medicating them." - yes, laziness is one of the problems. Also, the docs can earn more just passing out amphetamines left and right.
@DominicMichaelis The "buzz" of Ritalin is a bit different from amphetamines proper, but the overall chemical effect on the body is the same, so...
@J.M. actually methylphenidate has a paradox effect (i hope it is the right word) for guys with ADHD it helps concentrating but for healthy it affects not to be able to concentrate
@caveman Well, that would be due to the difference is neurotransmitter balances (in theory).
@DominicMichaelis That's true. Most of the other stimulants actually display the same paradox. As to why meth is not being given to kids, since it's cheaper... >:)
@DominicMichaelis Yes, that's a good analogy. :) Our understanding of the human body now in the 21st century is only slightly better, but still piss-poor.
It doesn't help that most of medical education these days is more "follow the diagnostic flowchart", without allowing most to think about why they do what they do.
I think most people don't appreciate that the human body is a complicated dynamical system. Mathematicians now understand that with a lot of dynamical systems, a tiny change in the initial conditions can lead to wildly different behavior. The medics, on the other hand, persist with their shotguns.
@DominicMichaelis this might be true, but I think one of the problems is the major part of people don't believe in themselves, but they have a great potential.
You can also look back and see what exactly you failed at. If it is easily fixed, then try again. If it is, say, a prerequisite which you do not know, then there is no point in trying again until you resolve that dependency.
@caveman This hugely depends on what effort you put in. Did you try everything, work hard, etc., or did you just freewheel along until a week before the exam? Also, looking at your corrected work can greatly help in understanding your thought processes and common mistakes. This has helped me do better on many exams (not just those which I failed).
@Chris'ssisterandpals, yeah - I heard this and it's motivating but I have trouble imagining what I will get from it (other than knowing a bunch of difficult math) if I don't pass my exams
I asked one of the staff at uni and he just said dont worry about it like .. I think they don't care about failures
@caveman: Having learned a good amount about a certain research area will both enable you to delve in deeper and to connect your new experience with other fields. This can greatly enhance the big picture.
@caveman: you don't seem to have the profile of someone that fails doing things. (I know a bit of psychology and I talked to you many times to figure it out). ;)
I have a question on defintion of derivative $\partial_X Y$ of vector field $Y$ in direction of tangent vector $X$. I was told that it is linear only with respect to $X$, but then I think that I don't understand definition
What is $\partial_X Y$ for vector field $Y$ on surface in $3d$ and for some tangent vector $X$? Is it a limit of increment of $Y$ "in direction X" divided by increment of something other?
but for surfaces in $\mathbb{R}^n$ it doesn't matter, some vector-valued smooth function of points on my surface such that vector in point $x$ is tangent to my surface.
I'm working on: Prove that $l\in(1,1.4)$, where $\displaystyle l=\sum_{k=2}^{\infty}\frac{1}{k (k!)^{1/k}}$. I couldn't find a nice way to prove the upper bound.
@Lord_Farin and if, say, I have local representation of my surface by radius-vector $r(u_1,\ldots,u_k)$ and $X(u) = \frac{\partial r}{\partial u_1}$, $Y(u)$ is some arbitrary vector field, do you know how to compute it by hands? Should I compute some increment of $Y(u)$?
Well, $TM$ is always "some $\Bbb R^n$" but without choosing coordinates. $TTM = TM$ always, because we can obviously make a path in $TM$ that has any vector in $TM$ as its tangent at zero.
Hey there all. Has this ever happened to any of you: someone going through old solutions you post, upvoting them, and then, some seconds later, removing the upvote? Sounds like a trifle, but it is infuriating while I am watching.
@Lord_Farin yes! very good! So if we have local parametrisation $r = r(u_1,\ldots,u_k)$, $Y(u)$ is a tangent vector field and $X$ is a tangent vector then by definition $\partial_X Y = \frac{\partial Y_i}{\partial u_j} X$, right?
@RonGordon Hmm. I'd be happy if I were you. I know a woman that was crashed in a street cross. Besides here car being wrecked, she is taking physical rehab. It sucks ass.
@Lord_Farin oh, yes, this seems intuitive, $i$-th coordinate of new vector is a directional derivative of $i$-th component of $V$ in direction $X$, just as we have in mathematical analysis for vectors-valued functions on $\mathbb{R}^n$
