Find limit of sequence $e^{-n}\sum_{k=0}^n n^k/k!$ I can't deal with finding: $$\lim_{n\to\infty}e^{-n}\sum_{k=0}^n n^k/k!$$ Have you any idea ?
Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ I'm supposed to calculate: $$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$$ By using WolframAlpha, I might guess that the limit is $\frac{1}{2}$, which is a pretty interesting and nice result. I wonder in which ways we may approach it.
So I was searching for one of my own websites on Google: http://www.botlibre.com  and to my surprise found some clones of my website, like:  http://83waystowasteyourtime.adidas.com  and:  http:// mediaiq.astral.comm  My question are:   Why are they creating clones of my website?  Some kind of SEO scheme or what? What gain could they be getting? How do they have a URL at adidas.com and astral.comm?  I assume it's not Adidas steeling my content. How do I shutdown these sites and stop people from stealing my content? (on Tomcat) In a perfect world, the so-important advice "do not reinvent the wheel" would be extensively followed. In practice, everyday millions of wheels are reinvented simply because someone didn't know an existing solution for a specific problem. The best way to find a library, today, is googling it, but it's not always obvious which keywords to use. The lack of a tool for finding solutions is very counterproductive for the dev world as a whole. Stack Overflow is not very receptive for recommendation threads. So what about a new SE for it? Convert HTML to PDF in .NET I want to generate a PDF by passing HTML contents to a function. I have made use of iTextSharp for this but it does not perform well when it encounters tables and the layout just gets messy.  Is there a better way? How to collapse consecutive numbers into ranges? I'm confused, does the absorption and emission determine the color of something? Or does that only happen when something is emitting energy?  When light hits an object, the photons get absorbed, then emitted with a different wavelength right? Why is the determinant the volume of a parallelepiped in any dimensions? For $n = 2$, I can visualize that the determinant $n \times n$ matrix is the area of the parallelograms by actually calculating the area by coordinates.  But how can one easily realize that it is true for any dimensions? How to disable Ctrl+Shift+U? Ctrl+Shift+U is a combined shortcut in Eclipse. But when I press it, it shows up a u, just like input with an underline. I guess this shortcut has been declared by Ubuntu, so I can't use it. I used to solve this problem by typing Caps Lock first. Is there any better method? How do we know the brain flips images projected on the retina back around? How do I see that every left ideal of a square matrix ring over a field is principal? As the question title suggests, how do I see that every left ideal of a square matrix ring over a field is principal? From the Cracks in time and space story arc it's known that if a TARDIS is blown up (possibly while in the Time Vortex) it rips creation itself apart. Also, from the Stolen Earth / Journey's End we learn the Daleks' masterplan which is subjugating creation itself via destroying it with the Reality Bomb (unfading Z-neutrino energy radiation).  In the Last Great Time War all sorts of temporal weapons were used and it's known that Daleks have the ability of time travel as well. Why didn't they use the Time Vortex back then to win the war once and for all via detonating a TARDIS while inside? I keep hearing about tandem breakers, and how they won't work with 240V loads and how they should never be used for a multi-wire branch circuit.  Aside from those issues...  What is a tandem breaker?     (note: they can be thrown independently). Term for group of guys attempting to show who is alpha? Similar to pissing contest - but not in an aggressive manner. Just each guy attempting to out talk one another to show they are the leader / in charge.   Sample sentence:     Well, that meeting was more of a X than being productive with Greg trying to show he was in control. The use of “So X as to be Y” First of all, note that this is not a duplicate question of another one asking about the usage of so as to, since this one is asking about the structure of so . . . as to. . . .  I understand this might have been a better question at the , since I’m not a native English speaker. I am asking it here since I want to get some professional advice in regard to this question, since I don’t seem to have found anything similar to this on the Internet.  At the Oxford Dictionaries Online (OD) I have found the following definition of ludicrous:     So foolish, unreasonable, or out of place as to be amusing.   I understand the phrase so as to, simply meaning in order to and preceding the purpose for which an action is done.  So by itself is just therefore, hence, thus, or a way of emphasizing an adjective or adverb.  As to in itself means with regard to or according to, as pointed out in .  In the OD definition of ludicrous, the use of so looks fair to me, that is, it clearly emphasizes the high extent to which something is foolish or unreasonable.  The use of as to doesn’t work in any way I’ve mentioned in this case (neither as in with regard to nor according to). I think the dictionary I’ve referred to has missed one, if not more, meaning of the phrase as to.  In this case, it appears to have likely been used as an alternative to that when used in the form So X, that Y, meaning something is or happens X (a combination of adjectives or adverbs) enough to cause Y to happen.  Of course, I don’t mean that in the structure so X as to Y the phrase as to is absolutely equivalent to that, since substituting as to with that changes the structure of Y.  Does all this mean that the phrase as to is a great alternative to that (although changing the structure of Y) whenever the sentence is of the form so X, that Y (where X is a combination of one or more adjectives or adverbs)?  I am new here and so I’m not too familiar with the variety of existing tags here, so if anyone sees an additional one that would fit, please add it to this question. I would also be glad to see some other corrections of my grammar. Revert new tab page behavior in Google Chrome 29+ I don’t like the behavior of the new tab page in Google Chrome 29+. How can I restore the old behavior?  The new tab page (if you have Google set as the default search engine) now shows the Google logo (or, apparently, the doodle of the day if there is one), a search box, and a few recently visited websites. It also added an Apps button to my favorites bar, seemingly the only way to access my apps page, which has easy access to things I use regularly (Calendar, Gmail, Docs, G+, Facebook, Feedly, Google News, Netflix, Play Music, Pandora, Drive, Keep, etc.) I don’t like this behavior. I actually don’t mind having the logo of the day and search box, but I want my apps linked to on my new tab page.  Is there any way to get the old behavior back? I checked the settings and it doesn’t seem like it’s possible to set chrome://apps to be the new tab page (which would restore the old behavior) or to configure the new new tab page to display apps instead of recently visited sites. jQuery doesn't work after content is loaded via AJAX I'm using 17.04 ( Clean Install ), and as far as I remember, every try to install any sofware- it fails with that error code.  I tried sudo apt-get update &amp;&amp; apt-get upgrade and problem persists.  any idea why?   edit: Using software Updater gave this result :   edit : output of update  guy@LG-LAPTOP:~$ sudo apt-get update &amp;&amp; apt-get upgrade [sudo] password for guy:  Ign:1 http://il.archive.ubuntu.com/ubuntu zesty InRelease Ign:2 http://il.archive.ubuntu.com/ubuntu zesty-updates InRelease               Ign:3 http://il.archive.ubuntu.com/ubuntu zesty-backports InRelease             Err:4 http://il.archive.ubuntu.com/ubuntu zesty Release                           503  Service Unavailable Get:5 http://security.ubuntu.com/ubuntu zesty-security InRelease [89.2 kB]      Err:6 http://il.archive.ubuntu.com/ubuntu zesty-updates Release                   503  Service Unavailable Get:7 http://archive.canonical.com/ubuntu zesty InRelease [10.2 kB]             Err:8 http://il.archive.ubuntu.com/ubuntu zesty-backports Release                 503  Service Unavailable Get:9 http://archive.canonical.com/ubuntu zesty/partner Sources [1,416 B]       Get:10 http://archive.canonical.com/ubuntu zesty/partner amd64 Packages [1,616 B] Get:11 http://archive.canonical.com/ubuntu zesty/partner i386 Packages [1,612 B] Get:12 http://archive.canonical.com/ubuntu zesty/partner Translation-en [708 B] Get:13 http://security.ubuntu.com/ubuntu zesty-security/main amd64 DEP-11 Metadata [11.7 kB] Get:14 http://security.ubuntu.com/ubuntu zesty-security/main DEP-11 64x64 Icons [9,219 B] Get:15 http://security.ubuntu.com/ubuntu zesty-security/universe amd64 DEP-11 Metadata [14.4 kB] Get:16 http://security.ubuntu.com/ubuntu zesty-security/universe DEP-11 64x64 Icons [31.0 kB] Reading package lists... Done                                          E: The repository 'http://il.archive.ubuntu.com/ubuntu zesty Release' does no longer have a Release file. N: Updating from such a repository can't be done securely, and is therefore disabled by default. N: See apt-secure(8) manpage for repository creation and user configuration details. E: The repository 'http://il.archive.ubuntu.com/ubuntu zesty-updates Release' does no longer have a Release file. N: Updating from such a repository can't be done securely, and is therefore disabled by default. N: See apt-secure(8) manpage for repository creation and user configuration details. E: The repository 'http://il.archive.ubuntu.com/ubuntu zesty-backports Release' does no longer have a Release file. N: Updating from such a repository can't be done securely, and is therefore disabled by default. N: See apt-secure(8) manpage for repository creation and user configuration details. guy@LG-LAPTOP:~$    output of apt get  guy@LG-LAPTOP:~$ sudo apt-get install putty Reading package lists... Done Building dependency tree        Reading state information... Done The following additional packages will be installed:   putty-tools Suggested packages:   putty-doc The following NEW packages will be installed:   putty putty-tools 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 661 kB of archives. After this operation, 2,713 kB of additional disk space will be used. Do you want to continue? [Y/n]  Err:1 http://il.archive.ubuntu.com/ubuntu zesty/universe amd64 putty-tools amd64 0.67-3   503  Service Unavailable Err:2 http://il.archive.ubuntu.com/ubuntu zesty/universe amd64 putty amd64 0.67-3   503  Service Unavailable E: Failed to fetch http://il.archive.ubuntu.com/ubuntu/pool/universe/p/putty/putty-tools_0.67-3_amd64.deb  503  Service Unavailable E: Failed to fetch http://il.archive.ubuntu.com/ubuntu/pool/universe/p/putty/putty_0.67-3_amd64.deb  503  Service Unavailable E: Unable to fetch some archives, maybe run apt-get update or try with --fix-missing? guy@LG-LAPTOP:~$ How can I send an email using a template? I have created a email template in ExactTarget API and I want to use it executing XML from SoapUI.  Is there any tutorial or examples of how can this be done?  I'm new at this and I'm not sure how ExactTarget really works. How to enable top bar background opacity? There is a way to change top bar background opacity in ubuntu 20.04 lts.  I installed the hide top bar extension but is not a option to change it. So the vending machine tore my $5 in pieces. What now? The vending machine swallowed half of my $5 bill, and the other half remained in my hand. Can I get my $5 back, or is it all lost forever? Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian. Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.   This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem to crack it.   (Please note that $e$ in the question is the group's identity.)  Here's my attempt though...  First I understand Abelian means that if $g_1$ and $g_2$ are elements of a group $G$ then they are Abelian if $g_1g_2=g_2g_1$...  So, I begin by trying to play around with the elements of the group based on their definition...  $$(g_2g_1)^r=e$$  $$(g_2g_1g_2g_2^{-1})^r=e$$ $$(g_2g_1g_2g_2^{-1}g_2g_1g_2g_2^{-1}...g_2g_1g_2g_2^{-1})=e$$  I assume that the $g_2^{-1}$'s and the $g_2$'s cancel out so that we end up with something like,  $$g_2(g_1g_2)^rg_2^{-1}=e$$ $$g_2^{-1}g_2(g_1g_2)^r=g_2^{-1}g_2$$  Then ultimately...  $$g_1g_2=e$$  I figure this is the answer. But I'm not totally sure. I always feel like I do too much in the pursuit of an answer when there's a simpler way.  Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra.