Does the Sum from N=1 to N=∞ of : (-1^(N+1))((1/P(N)), where P(N) is the nth prime number, converge? If so, to what number does it converge?  Or in different terms, does the sum of the reciprocal of primes, where there are alternating signs, converge?  Example: (1/2)-(1/3)+(1/5)-(1/7)+(1/11) . . . = ?
$$\sum_{n = 1}^\infty \frac{(-1)^n}{p_n}$$ where $p_n$ is the $n$th prime.  I have computed this to 10000 rather than infinity. My results suggest that convergence does happen but it's very slow. But I can't even be sure about the first few digits: $-0.26959$?  I have looked at the "questions that may already have your answer" and some of the "similar questions," but they involve somewhat different formulas.
2D graphics - why use spritesheets? I have seen many examples of how to render sprites from a spritesheet but I havent grasped why it is the most common way of dealing with sprites in 2d games.  I have started out with 2d sprite rendering in the few demo applications I've made by dealing with each animation frame for any given sprite type as its own texture - and this collection of textures is stored in a dictionary. This seems to work for me, and suits my workflow pretty well, as I tend to make my animations as gif/mng files and then extract the frames to individual pngs.  Is there a noticeable performance advantage to rendering from a single sheet rather than from individual textures? With modern hardware that is capable of drawing millions of polygons to the screen a hundred times a second, does it even matter for my 2d games which just deal with a few dozen 50x100px rectangles?  The implementation details of loading a texture into graphics memory and displaying it in XNA seems pretty abstracted. All I know is that textures are bound to the graphics device when they are loaded, then during the game loop, the textures get rendered in batches. So it's not clear to me whether my choice affects performance.  I suspect that there are some very good reasons most 2d game developers seem to be using them, I just don't understand why. How do I change the computer name? When I connect to my server (ubuntu server 10.10), I get this:  name@server-name.belkin ~&gt;   How can I remove ".belkin"? I lost the sa password on a machine, and when I log in to the machine directly using an account in the admin group, SQL Server Management Studio will not allow me to log in using Windows authentication.  My plan was to simply log into the server, connect via Windows Authentication, and reset sa to use a new password. Since I cant' connect via Windows Authentication, this won't work.  How else can I reset the sa password? I have been wondering why anyone ever thought that we could find an expansion for eigenstates of some perturbed Hamiltonian in terms fo those for the basis Hamiltonian. My lecturer insisted that this was because the eigenstates of the Hamiltonian are complete', however from my (limited) understanding of the mathematical basis of QM:   When you are given a Hamiltonian, it comes with the Hilbert space of functions of which it acts (subject to some conditions we impose, such as the Lebesgue square-integrability of functions) When you are given another Hamiltonian, there is no guarantee that it will act on the same Hilbert space Therefore, that the zeroth order Hamiltonian has eigenstates forming a complete basis for the space on which ti acts does not guarantee that these same eignestates form a complete basis for the Hilbert space on which the purturbed Hamiltonian acts.   As a simple example. I can imagine adding a small perturbing magnetic field to a Hamiltonian in which there initially is none, such that we lose the degeneracy in the eigenstates accounting for different spins of a particle. One could say that we have the same Hilbert space before and after but we lose degeneracy; or I would have been tempted to say that our initial Hilbert space did not have these extra dimensions corresponding to the spin states. So rather than the eigenstates of the basis Hamiltonian also forming a complete basis for the purtubed Hamiltonian, but having eigenvalue degeneracy which is lifted on the purtubration, I would have said that our Hilbert space has changed. After all, if we take the former view (i.e. a basis accouting for spin but with degeneracy) then it seems like for any problem we would need to know and account for EVERY possible system property that can be discerned. This doesn't seem reasonable to me....  NOTE:  1) My question was asked here previously:   although there is only a discussion and no conclusive answer given. I see that the distinction is important in relativistic QM, and the case of adding a potential onto a free particle Hamiltonian seems to be a special one. Although I am still not sure what the answer is for general purturbations.  2) Similar question here:   but it seems to completely avoid answering the question. I want to clear a element from a vector using the erase method. But the problem here is that the element is not guaranteed to occur only once in the vector. It may be present multiple times and I need to clear all of them. My code is something like this:  void erase(std::vector&lt;int&gt;&amp; myNumbers_in, int number_in) {     std::vector&lt;int&gt;::iterator iter = myNumbers_in.begin();     std::vector&lt;int&gt;::iterator endIter = myNumbers_in.end();     for(; iter != endIter; ++iter)     {         if(*iter == number_in)         {             myNumbers_in.erase(iter);         }     } }  int main(int argc, char* argv[]) {     std::vector&lt;int&gt; myNmbers;     for(int i = 0; i &lt; 2; ++i)     {         myNmbers.push_back(i);         myNmbers.push_back(i);     }      erase(myNmbers, 1);      return 0; }   This code obviously crashes because I am changing the end of the vector while iterating through it. What is the best way to achieve this? I.e. is there any way to do this without iterating through the vector multiple times or creating one more copy of the vector? Triggering actions when a variable is set Is it "5–6 weeks are a lot of time" or "5–6 weeks is a lot of time"? Sort Order on the User Filter is Surprising To Me When filling in the "Type to Find Users" , it seems to sort by exact match and then by reputation when I am on the reputation tab.   For example, please consider the following search results:    This threw me off since I thought that the sort order is reputation. I guess this often might give people what they want, but when picking a popular name is shows a lot of 0 rep users. Maybe adding a filter for low rep to the exact match might give more expected results, or just sorting on Rep? The bikeshed problem and Stack Exchange In short, :     If you go before the Board of   Directors and ask for 1.5 billion   dollars to build a nuclear reactor, no   one will review or discuss the details   of the plant. They will assume that   experts have been over every inch of   the plans, and not want to look   foolish by asking a silly question.      However, if you ask the same group to   approve a 30 dollar expenditure for   lumber with which to build a bikeshed   (presumably a British term for the   smallest possible building) then be   prepared for a 45 minute discussion   about all aspects of the bikeshed,   including the color of the paint. The   explanation for this is that everyone   can grasp the scale of a bikeshed.   To transform this problem to Stack Overflow terms, questions that are trivial such as "How to convert a C string to a QString or "What is the difference between . and ->" get a lot more views, answers and votes than real questions of the form "How do I frob this widget" that bother real professionals. Fewer views mean less possible votes for good answers and hence less incentive for people to answer.  How can Stack Exchange solve this problem? How do I create a script file for terminal commands? Finding conditional distribution Let $X$ and $Y$ be independent $Exp(1)$-distributed random variables. Find the conditional distribution of $X$ given that $X + Y = c$ ($c$ is a positive constant).  this is my idea:  $$f_{X \mid X+Y}(x\mid c) = \frac{f_{X,X+Y}(x,c)}{f_{X+Y}(c) } $$  if this is the right definition then i might calculate the righthand side to obtain the distribution but Iam unsure Stack Overflow provides good tools in the question editor, like "Links" and "Styling".  Another thing that for me would be nice is a color selector. Although an HTML editor is available, a direct click on tool would be good, and users can select different colors, as for marking errors in their program with red, and more important data with blue. Converting time field omitting time and returning null values? I have a time field in my shapefile that is formatted as HH:mm:ss.s It is a text field, and I want to convert it to a date so I can calculate time elapsed across rows (if there's a better way to do this let me know). Basically, each row is a point on a route, and I want to use the time field to calculate the amount of time elapsed between the first and last points on the route.   I tried using the convert time field tool, setting the input format as HH:mm:ss.s, and the output type as DATE. The output field however only shows values of 01/01/2001 for some rows and  for other rows.   time is the input field and time_c is the output field  I'm not sure what's going wrong here, here is what shows up in the geoprocessing results window:    I tried adding the date data to the field as well to make the format yyyy-MM-dd HH:mm:ss.s, but it still didn't work: How do I change Wi-Fi Hotspot settings in Ubuntu with GNOME 3? How do I manage my hotspot connected devices or change the hotspot password? I can only see the Network Name, Security type and the Password but I can't seem to manage users or the network properties. Can I fail to deserialize with XmlSerializer in C# if an element is not found? I am using XmlSerializer to write and read an object to xml in C#.  I currently use the attributes XmlElement and XmlIgnore to manipulate the serialization of the object.  If my xml file is missing an xml element that I require, my object still deserializes (xml -> object) just fine.   How do I indicate (preferably via Attributes) that a certain field is "required"?  Here is a sample method of what I am using currently:  [XmlElement(ElementName="numberOfWidgets")] public int NumberThatIsRequired {     set ...;     get ...; }   My ideal solution would be to add something like an XmlRequired attribute.     Also, is there a good reference for what Attributes are available to manipulate the behavior of XmlSerializer? Grandi's Series; tends to $1/2$, but why is this considered a valid sum? Grandi's series,  $$1+1-1+1-1+1-1+...$$  can be expressed as the below:  $$\sum_{n=0}^\infty(-1)^n$$  Two valid sums that make sense to me are $1$, and $0$, depending on how you approach the series. $(1+1)-(1+1)-(1+1)-...=0$, and $1+(1-1)+(1-1)+(1-1)+...=1$.  There is consensus, however, that the actual sum is $\frac{1}{2}$. Why? I understand the approach of finding partial means of the series, and they do indeed tend to $\frac{1}{2}$, but it seems unintuitive to assert that the sum is neither $1$ or $0$.  A more convincing method I found was assuming the series is $S$, then shifting it such that $S-1 = S$, then through algebra finding $S = \frac{1}{2}$, but again, it seems more intuitive answer is either $0$ or $1$. I say this strictly because adding and subtracting integers should equal an integer, never a fraction.  Is this a characteristic of infinite series, which is not specific to Grandi's series? Can I use the "Scan" button with SANE? I have an HP Deskjet with integrated scanner. It has a "Scan" button on its top, but this button does not interact with the xsane program on my Debian machine: To get a page scanned, I have to initiate the scanning process from xsane.  Is there a setting (or a scan program) that allows me to press the scan button and cause the program to scan a page? I want to prove with mathematical induction that:  $$\sum_{i=1}^n i \cdot i! = (n+1)! - 1$$  So in the first step we define $n = 1$:  $$\sum_{i=1}^1 i \cdot i! = 1 \cdot 1! = 1 = 2! - 1$$  In the second step we define $n = n + 1$:  $$\sum_{i=1}^{n+1} i \cdot i! = \sum_{i=1}^n i \cdot i! + (n+1)(n+1)!$$ $$\implies (n+1)! - 1 + (n+1)(n+1)!$$  This can now be simplified to $(n+2)(n+1)! - 1$ but I do not see how...  We could do $(n+1)!(-1 + (n+1)) = (n+1)! \cdot n$ but still this will not be the same as above.  What did I miss?  Bodo I'm interested in the properties of zero-diagonal symmetric (or hermitian) matrices, also known as hollow symmetric (or hermitian) matrices.   The only thing I can come up with is that it cannot be positive definite (if it's not the zero matrix): The Cholesky decomposition provides for positive definite matrices $A$ the existence of a lower triangular $L$ with $A=LL^*$. However the diagonal of $LL^*$ is the inner product of each of the rows of $L$ with itself. Since the diagonal of $A$ consists of zeros, so $L$ (and thus $A$) must be the zero matrix.  The sorts of questions that interest me are:   which symmetric matrices can be transformed orthogonally into a zero-diagonal matrix? what can we say about the eigen-values of a zero-diagonal symmetric matrix?. any other interesting known properties?? How to reduce AWS EBS root volume size?