Evaluate $$\sum_{n,k}  \frac{1}{\binom{n}{k}}, $$ where the summation ranges over all positive integers $n,k$ with $1&lt;k&lt;n-1$.  Thouhgts: We are trying to evaluate $$\sum_{n=4}^{\infty} \sum_{k=2}^{n-2} \binom{n}{k}^{-1}$$ We may try to find a closed form of the inner summation which is of the form : $$ \frac{1}{ {n \choose 2} } +  \frac{1}{{n \choose 3} }+ \dotsb + \frac{1}{{n \choose n-2} }.   $$ Notice that we may write $\frac{1}{ {n \choose 2} } = \frac{2! }{n(n-1)}$ and if keep doing the same for the other terms we obtain the following: $$ \frac{ (n-2)! + (n-3)! (n-(n-2)) + (n-4)!(n-(n-2))(n-(n-3)) + \dotsb + 2! (n-3)! }{n!}, $$ which equals $$ \frac{ (n-2)! + 2!(n-3)! + 3! (n-4)! + \dotsb + (n-3)! 2! }{n!} $$ and so this equals: $$ \frac{1}{n(n-1)} + \frac{2}{n(n-1)(n-2)} + \dfrac{6}{n(n-1)(n-2)} + \dotsb + \dfrac{2}{n(n-1)(n-2) }. $$ But half of this terms are identical. Therefore, we are trying to sum up series of the form $$\sum_{n \geq k} \frac{1}{(n-1)(n-2)(n-3)\dotso(n-k)} ,$$ which can be done by a telescoping trick, but it seems very formidable. Am I approaching this problem the right way? Any hints/suggestions?
How to calculate the sum of sequence $$\frac{1}{\binom{n}{1}}+\frac{1}{\binom{n}{2}}+\frac{1}{\binom{n}{3}}+\cdots+\frac{1}{\binom{n}{n}}=？$$ How about its limit?
Can a Dungeons and Dragons 4e character take D&D Essentials feats? In Heroes of the Fallen Lands many of the core feats have been streamlined, some paragon feats have become available to all levels.  Before that feats were tiered by level: heroic, paragon, epic.  Can a character that was made before the Heroes of the Fallen Lands publication use Heroes of the Fallen Lands feats? Logging in as the same user multiple times in Remote Desktop Windows Server 2008 Quick question: I have a situation where I need to let multiple people on different PCs log into one server 2008 machine as administrator simultaneously over remote desktop. I have the CALs for it, it's just not set up correctly. When one user tries to log in, it boots the other out. What I need is to present to them a different session, just each as logged in as admin. Sorry for the slightly rambling post, I'm new here. Thanks! Equivalent of cURL for Ruby? Create a variable in .CSS file for use within that .CSS file Possible Duplicate:          We have some "theme colors" that are reused in our CSS sheet.  Is there a way to set a variable and then reuse it?  E.g.  .css OurColor: Blue  H1 {   color:OurColor; } I would like to build a new computer (it's been 7 years since the last one). Does Windows 10 support more than one ? Proof: in $\mathbb{R}$, $((0,1),|\cdot|)$ is not compact. Let $(M,d)$ be a metric space, and $A\subset M$. By definition, $A$ is said to be compact if every open cover of $A$ contains a finite subcover.  What is wrong with saying that, in $\mathbb{R}$, if $I=(0,1)$, we can choose $G=\{(0,\frac{3}{4}), (\frac{1}{4}, 1)\}$, which satisfies $I \subset \bigcup_{U\in G} U$, but we can't extract a finite subcover, so $I$ is not compact. Is $G$ a finite subcover of $G$, so it is not a valid cover for proving this? I would take $\cup_{n\in\mathbb{N}} (\frac{1}{n},1)$ in order to prove this, can we conclude that every open cover is necessarily a infinite union of open sets $\neq \emptyset$? when does a functor map products into products? How can I create launcher icon for PhpStorm? How to know where shell variables and functions are set? When I type the set command in my system I've got this extract out :  __colormgr_commandlist='     create-device     create-profile     delete-device     delete-profile     device-add-profile     device-get-default-profile     device-get-profile-for-qualifier     device-inhibit     device-make-profile-default     device-set-kind     device-set-model     device-set-serial     device-set-vendor     find-device     find-device-by-property     find-profile     find-profile-by-filename     get-devices     get-devices-by-kind     get-profiles     get-sensor-reading     get-sensors     get-standard-space     profile-set-filename     profile-set-qualifier     sensor-lock     sensor-set-options     ' __grub_script_check_program=grub-script-check _backup_glob='@(#*#|*@(~|.@(bak|orig|rej|swp|dpkg*|rpm@(orig|new|save))))' _xspecs=([freeamp]="!*.@(mp3|ogg|pls|m3u)" [cdiff]="!*.@(dif?(f)|?(d)patch)?(.@([gx]z|bz2|lzma))" [bibtex]="!*.aux" [rgview]="*.@(o|so|so.!(conf|*/*)|a|[rs]pm|gif|jp?(e)g|mp3|mp?(e)g|avi|asf|ogg|class)" [oowriter]="!*.@(sxw|stw|sxg|sgl|doc?([mx])|dot?([mx])|rtf|txt|htm|html|?(f)odt|ott|odm)" [chromium-browser]="!*.@(?([xX]|[sS])[hH][tT][mM]?([lL]))" [tex]="!*.@(?(la)tex|texi|dtx|ins|ltx|dbj)" [netscape]="!*.@(?([xX]|[sS])[hH][tT][mM]?([lL]))" .../.. _xinetd_services ()  {      local xinetddir=/etc/xinetd.d;     if [[ -d $xinetddir ]]; then         local restore_nullglob=$(shopt -p nullglob);         shopt -s nullglob;         local -a svcs=($( printf '%s\n' $xinetddir/!($_backup_glob) ));         $restore_nullglob;         COMPREPLY+=($( compgen -W '${svcs[@]#$xinetddir/}' -- "$cur" ));     fi } dequote ()  {      eval printf %s "$1" 2&gt; /dev/null } quote ()  {      local quoted=${1//\'/\'\\\'\'};     printf "'%s'" "$quoted" } quote_readline ()  {      local quoted;     _quote_readline_by_ref "$1" ret;     printf %s "$ret" }   I checked all files in my knowledge as /etc/profile, /etc/environment, and/or ~/.bashrc. I didn't find any generation  script or code calling.   Do you have any advice where is come from ? SQL Services Wont Start After Removing TLS I just created a plane and added a wave modifier to it, to simply make it look like something fluid. Now, I have an Object that I want to place on the surface of this fluid plane. But how do I do that?  I thought of adding a driver to my Objects location, but I have no clue how to get the location of a certain vertex in runtime. Do indexes work with "IN" clause Dependency injection and Libraries/Frameworks $\require{enclose}\enclose{horizontalstrike}{\rm Greatest}\!$ Least prime factor of $n$ is less than square root of $n$, proof I remember reading this somewhere but I cannot locate the proof. Iptables rules with ! character I have entered iptables rules to log a specific connection on our server, but it doesn't seem to be working as I want it to. Here are the rules:  1    LOG        tcp  -- !10.51.0.0/16         0.0.0.0/0           state NEW    tcp dpt:8040 LOG flags 0 level 4 prefix `New Connection_8040TCP: ' 2    LOG        tcp  -- !192.168.0.0/16       0.0.0.0/0           state NEW tcp dpt:8040 LOG flags 0 level 4 prefix `New Connection_8040TCP: '   Iptables should log only IP connections which are not in the specified networks, but it keeps logging all connections from all IP addresses. Show that antipodal points remain antipodal under any isometry of $S^2$ "The antipodal map of $S^2$ is the isometry $\bar m$ sending each point $P$ to its opposite point $-P$" (from textbook)  Show that antipodal points remain antipodal under any isometry of $S^2$. In other words, for each isometry f of $S^2$, if $P$ and $P'$ are antipodal, then so are $f(P)$ and $f(P')$  I am unsure how I can define the $P$ and $P'$ so that they are opposites of each other in the antipodal map, and am I supposed to apply the traditional isometries (translation, reflection and rotation) to the points? I don't understand this problem. How do I unit test code that includes a callout? This is a canonical question and answer developed by the community to help address common questions. If you've been directed here, or your question has been closed as a duplicate, please look through the resources here and use them to shape more specific questions. To browse all canonical questions and answers, including more unit test resources, navigate to the  tag.  How do I unit test code that includes a SOAP or REST-based callout, or indirectly calls code that makes a callout? I'm getting an error saying that callouts are not supported. Lufthansa connecting flight at Munich, only 45 minutes in between. Will I be able to make it? Show that if $ \ A \cup B = A$ and $ \ A\cap B = A$ then $ \ A = B$ Question:  Show that if $ \ A \cup B = A$ and $ \ A\cap  B = A$ then $ \ A = B$  My attempt:  Proof by contradiction:  Assume $ \ A \cup B = A$ and $ \ A\cap  B = A$ and $ \ A \neq B$   Case 1: $ \exists \ x \in A, x\notin B$  If $ x \in A \implies x\in A \cap B \implies x \in A \ and \  x\in B  \implies x\in B$, a contradiction.   Case 2: $ \exists \ x \in B, x\notin A$  If $ \ x \in B \implies x\in A \cup B \implies x\in A$, since $ \ A \cup B = A$. Contradiction.   Is this approach correct? Could someone please show me how to do a direct proof? Recently, I read this  about how they have some of their infrastructure laid out. I was curious to know, does Stack Exchange use caching (both at the application layer and/or database layer) and if so, how?