So here's the deal, I gave a couple of my friends the password to my WIFI, but since then the internet has been going too slow for long periods of time.  I suspect that one of them is using a torrent application to download movies or something, I want to know who that person is by knowing which pc is downloading through torrent.  I have the admin privileges on my network adapter, and would also like to know if there is away to see who is torrenting though my network using only my laptop by any means. so please help me.
My internet connection is very slow. I suspect someone in the network is using a BitTorrent client which takes huge amount of bandwidth.  Is there a way to detect who is using my bandwidth (network sniffing tools etc)?  I should have pointed out that I have no access to the router, what I am looking for is a way of detecting someone's MAC/IP if she uses Bittorrent.
How to show the number of domains the disk is split into is $1+ C_n^2 + C_n^4$? Assuming there are n points on a circle, then there are $C_n^2$ chords. If any three chords don't intersect in circle. How to show the circle will be divided into $1+ C_n^2 + C_n^4$ domains?  Picture below is a example of 5 points on circle. Forgot to submit consumption tax (GST) declaration form upon exiting Tokyo Xcode 6.1 won't let me create a new project I'm running Xcode 6.1 on OS X Yosemite and I'm unable to create a new project, as you can see below it just says 'loading'.  I've removed the product and org name for the purpose of this image. I have tried uninstalling but that didn't work. I am new to Xcode and coding in general so I would appreciate it if any solution could be given in lay terms. Convergence of the series $ \sum_{n=1}^{+ \infty } \frac{1}{n \cdot \ln(n) \cdot (\ln(\ln(n)))^{a}} $ I have a problem with a convergence of the series $ \sum_{n=1}^{+ \infty } \frac{1}{n \cdot \ln(n) \cdot (\ln(\ln(n)))^{a}}$ depending on the parameter $a$. I know that  $\frac{1}{n} \rightarrow 0$, $\frac{1}{\ln(n)} \rightarrow 0$, but I don't know how to deal with $\frac{1}{\ln(\ln(n)))^{a}}$ and how to get to the final answer. So far I solved tasks with  $\frac{1}{n ^{a}}$ and I know that $a&gt;1  \Leftrightarrow \sum_{n=1}^{+ \infty }\frac{1}{n ^{a}}$ is convergent, but in this task I cannot use it. Can you help me? How can I determine whether a 2D Point is within a Polygon? Can Resharper be set to warn if IDisposable not handled correctly? Is there a setting in Resharper 4 (or even Visual Studio itself...) that forces a warning if I forget to wrap code in a using block, or ommit the proper Dispose call in a finally block? “All you have to do is read” vs. “All you have to do is to read” I realize that an image histogram is a graphical display of an images tonal distribution (i.e. horizontal darks to lights, vertical pixel distribution), but how does one really use it and why? I mean, can't you determine everything you need just by looking at the image? $\dfrac{pq}{r}+\dfrac{qr}{p}+\dfrac{pr}{q}\in\Bbb Z\,\Rightarrow\, \dfrac{pq}{r},\dfrac{qr}{p},\dfrac{pr}{q}\in\Bbb Z$ Let $p,q,r$ be integers such that the symmetric sum of fractions $\dfrac{pq}{r}+\dfrac{qr}{p}+\dfrac{pr}{q}$ is an integer. Prove that each of the numbers: $\dfrac{pq}{r},\dfrac{qr}{p},\dfrac{pr}{q}$ is an integer. How to do this? Someone posted but deleted soon. What does "No bootable medium found" mean in VirtualBox? I would like to take a picture of a street during the day to depict how much traffic and cars there are on the street.  Is there anyway to do this with a long exposure photograph?  I have a Nikon D90 and I set the f stop down to 1/16 (the lowest of my lens) and set the time to 2 seconds. Doing this, the exposure is almost saturated when taken around 5pm with the sun setting behind a tree. Doing this, the cars passing by were very hard to make out against the stationary background, being visible as a blur in the color of the car.  Perhaps this is not possible and I should just do this at night where the cars lights are much brighter than the background.  Please suggest how else I might depict the traffic in a photograph?  Are daytime long exposure photographs even possible? After enabling hibernate grub menu appears on start-up with 30s timeout My system is not dual boot just single OS UBUNTU 16.04. Prior to making the following changes, the Grub screen would not appear at start-up (OK).  After following these steps from  to enable hibernate in the system menu:   Open terminal and enter;   sudo nano /etc/polkit-1/localauthority/50-local.d/com.ubuntu.enable-hibernate.pkla  add the lines into the empty file;  [Re-enable hibernate by default in upower] Identity=unix-user:* Action=org.freedesktop.upower.hibernate ResultActive=yes  [Re-enable hibernate by default in logind] Identity=unix-user:* Action=org.freedesktop.login1.hibernate;org.freedesktop.login1.handle-hibernate-key;org.freedesktop.login1;org.freedesktop.login1.hibernate-multiple-sessions;org.freedesktop.login1.hibernate-ignore-inhibit ResultActive=yes  ^X then Y to save (a filename was suggested and I accepted without taking note).  Reboot for the changes to take effect.   hibernate now works and appears in the menu (OK) but whenever I reboot I now get a Grub menu at start up with a full 30 second timeout (NOK). I tried playing around with the Grub menu timeout settings but still get the full 30 sec timeout.  How do I stop the Grub menu from appearing and reduce my boot time to what it was before I made these changes.   Current Grub settings;   GRUB_DEFAULT="0" GRUB_HIDDEN_TIMEOUT="0" GRUB_HIDDEN_TIMEOUT_QUIET="true" GRUB_TIMEOUT="1" GRUB_DISTRIBUTOR="`lsb_release -i -s 2&gt; /dev/null || echo Debian`" GRUB_CMDLINE_LINUX_DEFAULT="quiet splash" GRUB_CMDLINE_LINUX="" Fell for fake sugar daddy. How do I get out? I recently printed a paper that asks to prove the "amazing" claim that for all $a_1,a_2,\dots$  $$\sum_{k=1}^\infty\frac{a_1a_2\cdots a_{k-1}}{(x+a_1)\cdots(x+a_k)}=\frac{1}{x}$$  and thus (probably) that  $$\zeta(3)=\frac{5}{2}\sum_{n=1}^\infty {2n\choose n}^{-1}\frac{(-1)^{n-1}}{n^3}$$  Since the paper gives no information on $a_n$, should it be possible to prove that the relation holds for any "context-reasonable" $a_1$? For example, letting $a_n=1$ gives  $$\sum_{k=1}^\infty\frac{1}{(x+1)^k}=\frac{1}{x}$$  which is true.  The article is Remotely run script on Unix, get output locally? I need to run a (Tcl, or whatever) script on a remote Unix machine, from a (local) Windows command shell, and I want the stdout/stderr of the script to show up in the Windows CLI. The exit code would also be useful.  Is this possible using ssh (putty)? Or by any other means?  Thanks! Network reputation graph becomes nervous when clicking quickly, backs away slowly How do you disable an application specific hotkey, for example, "Command-P"? In many applications, Command-P is used to print.  My fingers are accustomed to emacs settings and reflexively press Command-P not for printing but for browsing history. It's annoying to see the print pop-up window every time when I press Command-P. I don't need a shortcut to print. How do you block a shortcut globally? I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks! Is there an alternative way to express "has them " other than "makes them "? For example, in the following sentence, can I replace have them with makes them, or is there another verb that more correctly I could use instead of has?     Since the manufacturers are so deeply involved, in some cases Wal-Mart even has them handle their own distribution, saving the retailer big bucks and increasing profit margins on cheaper goods.   Another example could be the following one.     The doctor sends my father and Uncle Johnnie out for all the ice they can get. Then he has them dump the ice in the bathtub, add cold water almost to the top, whereupon my aunt is wrestled into the tub and submerged. To ensure that a new chapter starts on a right-hand page, the book class (and the report class with the options twoside and openright) will insert blank pages between chapters if necessary. However, these pages still contain the headers and footers as defined by the class. How can I remove those?  Note: I know that I can manually add \clearpage\thispagestyle{empty} at the appropriate points of my document. However, this seems awkward and is prone to fail if I change my document later. I'm looking for automatic solutions.