I want to prove that $$\frac{c^n}{n!}$$ tends to zero when $n$, a positive integer, is quite large. I do not think I can use L'Hospital's rule here as $n$ is discrete. Should we assume $n$ to be continuous to sove this problem. Furthermore, I do not find approximating $n!$ with gamma function of much help.
Why is     $$\lim_{n \to \infty} \frac{2^n}{n!}=0\text{ ?}$$   Can we generalize it to any exponent $x \in \Bbb R$? This is to say, is     $$\lim_{n \to \infty} \frac{x^n}{n!}=0\text{ ?}$$     This is being repurposed in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: .
Topological/Geometrical justification for $\text{CFT}_2$ being special It is known as a fact that conformal maps on $\mathbb{R}^n \rightarrow \mathbb{R}^n$ for $n&gt;2$ are rotations, dilations, translations, and special transformations while conformal maps for $n=2$ are from a much wider class of maps, holomorphic/antiholomorphic maps. I was wondering to know if there is any topological or geometrical description for this.   To show what I mean, consider this example: in $\mathbb{R}^n$ for $n&gt;2$ interchanging particles can only change the wave function to itself or its minus. It is related to the fundamental group of $\mathbb{R}^n-x_0$ ($x_0$ is a point in $\mathbb{R}^n$ and $\pi_1(\mathbb{R}^n-\{x_0\})=e$ for $n&gt;2$) but this is not true for $n=2$.   I want to know whether exists any topological invariant or just any geometrical explanation that is related to the fact that I mentioned about conformal maps on $\mathbb{R}^n$. For equal cost, why would you select the MacBook Air over the MacBook Pro? I can see that the 13" version of MacBook Pro and MacBook Air is same $1199, but the Pro is more feature-loaded than Macbook Air. Why would anyone buy MacBook Air? What's the catch in this?    I was thinking of buying a MacBook for development purposes, and worried if there's any problem in the Pro version, because it comes with more features (like more storage).  Edit:    Thanks for the answer guys, I bought a MacBook Pro a week ago and already loving it. What does "very approximate" mean? An approximate answer is one which is close to the correct answer.  Likewise, we can talk of an approximate model, or approximate methods in mathematics.  The etymology is from the Latin ad, "to" and proximus, "closest", see e.g. .  I don't think this can be disputed.  Occasionally proximate is used as a synonym, though that word has a different meaning in legal English.  So if an approximate answer is close, a very approximate answer should be very close.  In practice, "very approximate" or "very approximately" is overwhelmingly used with the opposite meaning of "roughly", indicating that an answer or estimate is not necessarily accurate.  For examples,   Is there any justification for this use? Is there any justification for my suggested use?  Should we avoid the phrase altogether? When would I need a SecureString in .NET? How to use paging in a WFS query? I am running a GetFeature request against a WFS server which does not support to download all data at once. Can I use ogc:PropertyIsGreaterThanOrEqualTo to split the dataset into chunks at download them step by step? I wonder how I can construct the actual URL to include the paging filter. Here is what I tried:  http://example.com/wfs.aspx?request=GetFeature&amp;service=WFS&amp;version=1.1.0 \ &amp;typeName=example:example&amp;maxFeatures=50000 \ &amp;FILTER=&lt;ogc:Filter xmlns:ogc="http://www.opengis.net/ogc"&gt; \ &lt;ogc:PropertyIsGreaterThanOrEqualTo&gt;&lt;ogc:PropertyName&gt;OBJECTID&lt;/ogc:PropertyName&gt; \ &lt;ogc:Literal&gt;50000&lt;/ogc:Literal&gt;&lt;/ogc:PropertyIsGreaterThanOrEqualTo&gt;&lt;/ogc:Filter&gt;   This is not a valid URI though... That's why uri-encoded the filter part:  http://example.com/wfs.aspx?request=GetFeature&amp;service=WFS&amp;version=1.1.0 \ &amp;typeName=example:example&amp;maxFeatures=50000 \     &amp;FILTER=%3Cogc%3AFilter%20xmlns%3Aogc%3D%22http%3A%2F%2Fwww.opengis.net \ %2Fogc%22%3E%3Cogc%3APropertyIsGreaterThanOrEqualTo%3E%3Cogc%3APropertyName%3EOBJECTID \ %3C%2Fogc%3APropertyName%3E%3Cogc%3ALiteral%3E50000%3C%2Fogc%3ALiteral \ %3E%3C%2Fogc%3APropertyIsGreaterThanOrEqualTo%3E%3C%2Fogc%3AFilter%3E   This actually works. Is there a "nicer" way to do this? When should I add another answer? I have been member of SO for 4 months. I have a question regarding some functionality. Whenever I give an answer to a question, there is functionality given Add Another Answer. I can comment on my answer at any time and also able to edit answer. Then   Why this functionality given on SO to add another answer?  When should I add another answer?  Why there is requirement of multiple answers? What is limit of adding answers to single question? Activity restart on rotation Android How to create a site in a sitecollection on a remote SPServer using C# How to store and preserve lots of data? I have many DVDs with movies and games, they are getting old and dusty, some of them are damaged. And I have more than 300 GB of general files in my PC. I want to store and preserve all of my data and keep them for a long long time.  I want to know what is the best and more reliable storage media that I should use and... I can buy. I will copy all of my DVDs to a new media and throw all of my DVDs away. It will take a very smaller physical space.  I think that everything that I need to store is currently 800 GB.  I do not like optical discs, because they are fragile and I have lost many CDs and DVDs. Hard drives would be good, but they can suddenly stop working, and it is very expensive to recover data from a broken hard drive.  Other questions: How often will I have to copy the files to a new storage media? Is it good idea to have another copy of each file stored in another storage media?  Edit: If I put all of the movies and games in a HDD, if the HDD stops working, I will lose everything. This is how valuable they are for me: If I lose 10% of them, I will not cry, because I can get new ones, but if I lose 90%, it will be a disaster.  Edit 2: I do not think that that online storage is reliable even if I pay for it, because they may be closed or may get bankrupt. They are not from my country and it would be difficult to get the files back. How to upload string as file with jQuery or other js framework Let $f$ be a continuous and open map from $\mathbb R $ to $\mathbb R$.Prove that $f$ is monotonic. Let $f$ be  a continuous and open map from $\mathbb R $ to $\mathbb R$.Prove that $f$ is monotonic.  Suppose that $f$ is not monotonic.Then $\exists a,b;a&gt;b$ such that $f(a)&lt;f(b)$ and $c,d;c&lt;d$ such that $f(c)&gt;f(d)$.Since $f$ is continuous then there exists no break in the graph of $f$. But how should I use the fact that $f$ is open.Any help on how should I do the proof? I used the jQuery Form plugin for asynchronous form submission. For forms that contain files, it copies the form to a hidden iframe, submits it, and copies back the iframe's contents. The problem is that I can't figure out how to find what HTTP status code was returned by the server. For example, if the server returns 404, the data from the iframe will be copied as normal and treated as a regular response.  I've tried poking around in the iframe objects looking for some sort of status_code attribute, but haven't been able to find anything like that.    The $.ajax() function can't be used, because it does not support uploading files. The only way to asynchronously upload files that I know of is using the hidden iframe method. With biblatex, is it possible to replace one author of an author list by a bibnamedash or idem string? With biblatex, the author can be replaced by a dash (or “idem”) in the bibliography. Is it possible to have this replacement also if there are multiple authors, and the first of these multiple authors is the same as the author of the book listed before? (It is somewhat difficult to describe, see the attached example instead, where “Buchautor, Hans-Wilhelm” should be replaced in book3 as well.)   I assume that it should be possible by customising \DeclareNameFormat, but I don’t know how.  Minimal example:  \documentclass{scrartcl} \usepackage[T1]{fontenc} \usepackage[latin1]{inputenc} \usepackage[english]{babel} \usepackage{csquotes} \usepackage{filecontents} \begin{filecontents}{\jobname.bib} @BOOK{book1,   author = {Buchautor, Hans-Wilhelm},   title = {Irgendein Buch},   location = {Buch am Wald},   date = {2000} } @BOOK{book2,   author = {Buchautor, Hans-Wilhelm},   title = {Ein anderes Buch},   location = {Berlin},   date = {2001} } @BOOK{book3,   author = {Buchautor, Hans-Wilhelm and Zweitautor, Zoran},   title = {Noch ein Buch},   location = {London and New York},   date = {2010} } \end{filecontents}  \usepackage[ style=authortitle, backend=bibtex8 ]{biblatex} \bibliography{\jobname}  \renewcommand*{\bibnamedash}{%   \bibstring[\mkbibbold]{idem\thefield{gender}}}  \begin{document} \nocite{*}  \printbibliography \end{document}   Edit 15/02/2011 17:14   I should have added that there was a feature request for biber which pointed back to biblatex. See .  Edit 21/06/2011 9:26  Is this somehow possible now with biblatex 1.4/1.5? I played a bit with uniquename/uniquelist, but I could not get it to work. Where did the "Magic Blood" of the Harry Potter Universe Originate From? What is a wizard that makes them non-Muggle? Or, more directly, what ancestry does the witch/wizard gene follow? Is there a patriarchal wizard (or matriarchal witch) that Muggles have no part in? Elves perhaps? Stop reading here.  Purpose of the Question/History/Don't Read There seem to be solid consequences associated with being Muggle, or Muggle-related. A dominant theme in the novels is discrimination against the non-gifted race, but, when read into, the wizard bloodline doesn't appear all that well defined. JKR indicates that wizard-ness is genetic, or inherited:  &quot;Muggle-borns will have a witch or wizard somewhere on their family tree, in some cases many, many generations back. The gene resurfaces in some unexpected places.&quot;  By the above, it stands that a family that produces a squib (non-magical born of magical parents) must be a half-blooded family in some way, since half-blooded only requires one non-magical ancestor. Of course, we are dealing in magic genetics. Any other form of genetics would make this inherited, seemingly recessive trait (or squibs couldn't exist), never appear at all. Unless they were XMen. If it is so important, what does the wizard-bloodline follow? Other Mythos In other literature a wizard can be nearly anything, but magical powers are typically inherited as well as learned. Merlin of the Arthurian Legend (also mentioned in HP) is Cambian, meaning he is born of demonic intervention; this explains his source of enchantment. Gandalf of Middle Earth is a Maiar, as are all Wizards of Middle Earth. Wizards in Harry Potter seem to follow a bloodline, and it has been mentioned (in the question Gabe linked) that genetics have a heavy play in it, meaning that squibs and parents of Muggle-born wizards should have identical genetics. This bloodline must come from somewhere. Magical Creatures In Harry Potter it is common for magical creatures to &quot;mingle&quot; with humans, these beings are typically somewhat magical. Elves have been described to have very strong magic, but are for some unreasonable reason subservient to humans. I may be mistaken, but believe it was hinted that elves and the wizarding lineage had intersected at one time. Maybe I'm imagining this. Genocidal Allusions Voldemort hates muggles, and half-bloods, and whatever else is non-wizard. The presence of this theme, suggests that this wizardly bloodline must be at least defined if the main antagonist is so obsessed with keeping that mysterious bloodline pure. Since Slytherin himself was said to be a supremacist (circa a really long time ago) it suggests that this bloodline should have existed long before he was around, especially if he founded a school with people he was not related to, but were also magical. The adverb "quite" has the following meanings :     1: wholly, completely ("not quite   finished")      2: to an extreme : positively "quite   sure" —often used as an intensifier   with a "quite a swell guy" "quite a   beauty"      3: to a considerable extent : rather   ("quite near")   My understanding is that #1 and #2 conflict with #3. For example, when you say "She's quite right", do you mean "She's completely right" or "To a considerable extent, she's right"? Notice that in the latter case you're not saying that she is completely right.  Another example is "I'm quite sure". Does it necessarily mean "I'm completely sure"?   Does the word really have conflicting meanings that can possibly cause ambiguity? Or am I missing something? How to run my own program without specifying its path Let's suppose I have compiled something and I run it like so:  $ /path/to/my/executable/mycmd Hello World   What do I need to do to run it like   $ mycmd Hello World    from everywhere in my computer? When should I use "a" vs "an"? In the following example, is it appropriate to use a or an as the indefinite article, and why?     He ate __ green apple.   I know that in the case of just "apple", it would be "an apple," but I've heard conflicting answers for "green apple," where the noun is separated from the article by an adjective.  Also, which is more appropriate in this case:     He ate  __ enormous Pop-Tart. I was granted a visa to Australia few months ago. I’d like to travel to Australia in few days. However, I’ve just renewed my passport and I don’t have enough time to inform them about my passport renewal. Can I travel to Australia by carrying both my old and my new passports? I want to implement the EM algorithm manually and then compare it to the results of the normalmixEM of mixtools package. Of course, I would be happy if they both lead to the same results. The main reference is Geoffrey McLachlan (2000), .  I have a mixture density of two Gaussians, in general form, the log-likelihood is given by (McLachlan page 48):      $$ \log L_c(\Psi) = \sum_{i=1}^g \sum_{j=1}^n z_{ij}\{\log \pi_i + \log f_i(y_i;\theta_i)\}. $$ The $z_{ij}$ are $1$, if the observation was from the $i$th component density, otherwise $0$. The $f_i$ is the density of the normal distribution. The $\pi$ is the mixture proportion, so $\pi_1$ is the probability, that an observation is from the first Gaussian distribution and $\pi_2$ is the probability, that an observation is from the second Gaussian distribution.   The E step is now, calculation of the conditional expectation:      $$ Q(\Psi;\Psi^{(0)}) = E_{\Psi(0)}\{\log L_c(|\Psi)|y\}. $$ which leads, after a few derivations to the result (page 49):      \begin{align} \tau_i(y_j;\Psi^{(k)}) &amp;= \frac{\pi_i^{(k)}f_i(y_j;\theta_i^{(k)}}{f(y_j;\Psi^{(k)}}  \\[8pt]                        &amp;= \frac{\pi_i^{(k)}f_i(y_j;\theta_i^{(k)}}{\sum_{h=1}^g \pi_h^{(k)}f_h(y_j;\theta_h^{(k)})} \end{align} in the case of two Gaussians (page 82):      $$ \tau_i(y_j;\Psi) = \frac{\pi_i \phi(y_j;\mu_i,\Sigma_i)}{\sum_{h=1}^g \pi_h\phi(y_j; \mu_h,\Sigma_h)} $$ The M step is now the maximization of Q (page 49):      $$ Q(\Psi;\Psi^{(k)}) = \sum_{i=1}^g\sum_{j=1}^n\tau_i(y_j;\Psi^{(k)})\{\log \pi_i + \log f_i(y_j;\theta_i)\}. $$ This leads to (in the case of two Gaussians) (page 82):      \begin{align} \mu_i^{(k+1)} &amp;= \frac{\sum_{j=1}^n \tau_{ij}^{(k)}y_j}{\sum_{j=1}^n \tau_{ij}^{(k)}}  \\[8pt] \Sigma_i^{(k+1)} &amp;= \frac{\sum_{j=1}^n \tau_{ij}^{(k)}(y_j - \mu_i^{(k+1)})(y_j - \mu_i^{(k+1)})^T}{\sum_{j=1}^n \tau_{ij}^{(k)}} \end{align} and we know that (p. 50)      $$ \pi_i^{(k+1)} = \frac{\sum_{j=1}^n \tau_i(y_j;\Psi^{(k)})}{n}\qquad (i = 1, \ldots, g). $$ We repeat the E, M steps until $L(\Psi^{(k+1)})-L(\Psi^{(k)})$ is small.        I tried to write a R code (data can be found ).    # EM algorithm manually # dat is the data  # initial values pi1       &lt;-  0.5 pi2       &lt;-  0.5 mu1       &lt;- -0.01 mu2       &lt;-  0.01 sigma1    &lt;-  0.01 sigma2    &lt;-  0.02 loglik[1] &lt;-  0 loglik[2] &lt;- sum(pi1*(log(pi1) + log(dnorm(dat,mu1,sigma1)))) +               sum(pi2*(log(pi2) + log(dnorm(dat,mu2,sigma2))))  tau1 &lt;- 0 tau2 &lt;- 0 k    &lt;- 1  # loop while(abs(loglik[k+1]-loglik[k]) &gt;= 0.00001) {    # E step   tau1 &lt;- pi1*dnorm(dat,mean=mu1,sd=sigma1)/(pi1*dnorm(x,mean=mu1,sd=sigma1) +            pi2*dnorm(dat,mean=mu2,sd=sigma2))   tau2 &lt;- pi2*dnorm(dat,mean=mu2,sd=sigma2)/(pi1*dnorm(x,mean=mu1,sd=sigma1) +            pi2*dnorm(dat,mean=mu2,sd=sigma2))    # M step   pi1 &lt;- sum(tau1)/length(dat)   pi2 &lt;- sum(tau2)/length(dat)    mu1 &lt;- sum(tau1*x)/sum(tau1)   mu2 &lt;- sum(tau2*x)/sum(tau2)    sigma1 &lt;- sum(tau1*(x-mu1)^2)/sum(tau1)   sigma2 &lt;- sum(tau2*(x-mu2)^2)/sum(tau2)    loglik[k] &lt;- sum(tau1*(log(pi1) + log(dnorm(x,mu1,sigma1)))) +                 sum(tau2*(log(pi2) + log(dnorm(x,mu2,sigma2))))   k         &lt;- k+1 }   # compare library(mixtools) gm &lt;- normalmixEM(x, k=2, lambda=c(0.5,0.5), mu=c(-0.01,0.01), sigma=c(0.01,0.02)) gm$lambda gm$mu gm$sigma  gm$loglik   The algorithm is not working, since some observations have the likelihood of zero and the log of this is -Inf. Where is my mistake? Evaluate $\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)$ Evaluate $$\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)$$   I tried to create a telescoping sum but I couldn't. The last step I could reach was turning the limit into $$\lim_{n\to \infty} \sum_{r=1}^n \left(\frac {1}{2^r(1-\tan (2^{-r+1})} -\frac {1}{2^r(1+\tan (2^{-r+1})}\right) $$  But couldn't proceed further. Also I thought about Riemann sums but it was a pure dead end.  Any help would be greatly appreciated