Prove that $\sum \limits_{n=2}^{\infty}{\dfrac{1}{n\ln (n)}}$ diverge  I know it's a well-known series, but all the proofs I've seen are based on the integral test and Cauchy condensation test. I need to prove it using only the following tests: Direct/limit Comparison, Root, D'Alembert, Leibniz, since they are all I have studied so far. Regards and thanks a lot.
Study the convergence of the following series: $$\sum_{n=2}^\infty \frac{1}{n^\alpha \cdot\ln^\beta(n)} \text{ where }\alpha,\beta \geq 0 $$   Applying d'Alembert criterion I have that $$ \lim_{n\to\infty} \frac{n^\alpha  \ln^\beta(n)}{(n+1)^\alpha  \ln^\beta (n+1)} = \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^\alpha\left(\frac{\ln(n)}{\ln (n+1)}\right)^\beta = 1$$ so the nature of the series is inconclusive.  If $\alpha = \beta = 0$, then the series diverges, since $\sum_{n=2}^\infty 1 = \infty$.  Should I study the rest of the cases (i.e. if $\alpha = 0, \beta &gt; 0$ the root test and the ratio test are also inconclusive). What is the best form to study the series?.  Thanks in advance
I thought it was The Tommyknockers by Stephen King, but after a check with Wikipedia, it appears that's not the case.  What I remember about the story (a TV movie I think, but definitely based on a book) was that the world is created anew every night, and that the old, used up, Earth was eaten nightly by some scary creatures.  The major plot point involved some protagonists not getting onto the new world at midnight, and having to escape from the monsters that were eating up yesterday's planet.  Does this ring a bell for anyone? I'm finding it very hard to Google for. Can a shadow move faster than the speed of light? As title,  Is any finite-dimensional extension of a field, say $F$, algebraic and finitely generated?  Say if $K/F$ is a finite extension when $K$ is a finite-dimensional vector space of $F$. Clearly, this implies that $K$ is finitely generated (as an algebra) over $F$, since a basis is a generating set. So every finite extension is finitely generated.  So indeed they all are, is my logic correct? Is "You are correct" correct? Following a global preference for , as of now, the preferences for enabling keyboard shortcuts and disabling the fixed topbar are network-wide* as well.  If you had either checked off on any site in the network, that setting has been preserved and applied to the whole network for your account already. Future changes will also apply to all* Q&amp;A sites.  Find them in their usual spot as previously on the  pages.    * With the exception of , which will support the fixed topbar after the . Specifing ip address for outbound connections on a multi ip host one of my servers (Debian 5.0.6) has two public ip-address on the same interface. This used to work well for months but suddenly it is using "the wrong" ip addresses for outgoing connections. This is a problem because the reverse lookup will not match and emails therefore get spam points.  eth0      Link encap:Ethernet  Hardware Adresse 00:1b:21:14:8e:9c             inet Adresse:81.169.180.51  Bcast:81.169.180.51  Maske:255.255.255.255           inet6-Adresse: fe80::21b:21ff:fe14:8e9c/64 Gültigkeitsbereich:Verbindung  eth0:0    Link encap:Ethernet  Hardware Adresse 00:1b:21:14:8e:9c             inet Adresse:85.214.157.120  Bcast:85.214.157.120  Maske:255.255.255.255   Kernel-IP-Routentabelle Destination     Router          Genmask         Flags Metric Ref    Use Iface 81.169.180.1    0.0.0.0         255.255.255.255 UH    0      0        0 eth0 0.0.0.0         81.169.180.1    0.0.0.0         UG    0      0        0 eth0   Currently it is using 85.214.157.120 for outbound connections. How do I get it to use 81.169.180.51?  Edit: The netmask of 255.255.255.255 is consistent with both documentation and DHCP response of the hosting company. Calling /etc/init.d/networking restart multiple times will eventually end up with the correct ip-address for outbount connections. But that is obviously not a stable solution. /Edit  Edit 2: To make sure that the host route is not related to my issue, I setup a local test network:  eth0      inet Adresse:192.168.0.2  Bcast:192.168.0.255  Maske:255.255.255.0 eth0:0    inet Adresse:192.168.0.3  Bcast:192.168.0.255  Maske:255.255.255.0  192.168.0.0    0.0.0.0         255.255.255.0   U     0      0        0 eth0 0.0.0.0        192.168.0.1     0.0.0.0         UG    0      0        0 eth0   If anyone has an idea how to make sure that source ip-address 192.168.0.2 is used on outbound tcp connections, I'd be thankful. /Edit 2 Show the post's comments on the "Suggested Edits" review interface Cannot Buy Land in Falkreath Is there even an automatic ban on meta? Somewhere I heard the automatic question ban on Meta is set lower than on the main site. But looking at the profile of  it looks like there is absolutely no automatic ban on meta.  Is this intended behaviour? Dealing with "java.lang.OutOfMemoryError: PermGen space" error How do I get Gmail to ask me to be a mailto handler again? Heartbleed: What is it and what are options to mitigate it? Get the row and column name of the minimum element of a matrix Can we start cleaning up [table] tag and burninate it ASAP Related questions:     (This led to founding of )   The tag  has become cluttered with all kinds of questions ranging from sql tables (for all DBM systems, HTML tag &lt;table&gt;, outputting content in any language (PHP, C#, LaTeX to name a few) as a table and several plugins of various languages and/or IDEs (like jTable, tablesorters and the kind).  The current count stands at 18,315 questions having this ambiguous tag. A few of possible suggestions for cleaning process:    +  to  |   +  to  |   +  to  |   +  to  |   + other-specific-DBM systems to whatever_DBM_it_is-table  +  to  |    Since I've no idea about other languages and their specific features regarding table, I'd ask the community for their input on this (such as , ,  and ). The other questions are either tagged with arrays or sql and similar tags, and in my opinion, just removing  from such questions would be enough! At the very end of Deathly Hallows, after Harry repaired his own wand with the Elder Wand, he tells Dumbledore's Portrait in the office:     “I’m putting the Elder Wand,” he told Dumbledore, who was watching him with    enormous affection and admiration, “back where it came from. It can stay there. If I die a    natural death like Ignotus, its power will be broken, won’t it? The previous master will    never have been defeated. That’ll be the end of it.   Now, this seems incredibly foolish to me.  "The previous master will never have been defeated" would only have been assuredely true if Harry had died right after depositing the Elder Wand back into Dumbledore's Tomb.   If not, Harry (the Wand owner) would still be alive, and at the risk of being Disarmed (as was shown with Harry and Draco Malfoy, it doesn't matter if the Elder Wand's owner is defeated by death or simple disarming, AND it's not even required that the Disarming would involve the Elder Wand!).  So, if Harry merely put the Elder Wand back into the Tomb, and some years later, someone would disarm him (or, even worse, kill him) - then that someone would become the Elder Wand owner, either by accident or by design:    By accident - Harry is likely to get into plenty of magical duels in the future. Even discounting friendly one (teaching people the Expelliramus), he plans on becoming an Auror!  By design - plenty of people have been in the Great Hall hearing Harry explaining to Voldemort that he, Harry, was now the owner of the Elder Wand, so the fact of his ownership would most assuredly NOT remain a secret, inviting countless people tempted by owning the Deathstick to hunt Harry down. Oh, and the location is now also known!   So "its power will be broken" is quite unlikely to come true!  Is Harry hopelessly wrong? Or am I missing something in the above logic? Was this addressed in canon/JKR info? I have a question related to modeling short time-series. It is not a question , but how. What method would you recommend for modeling (very) short time-series (say of length $T \leq 20$)? By "best" I mean here the most robust one, that is the least prone to errors due the fact of limited numbers of observations. With short series single observations could influence the forecast, so the method should provide a cautious estimate of errors and possible variability connected to the forecast. I am generally interested in univariate time-series but it would be also interesting to know about other methods. Floating point comparison I often have to sort a dictionary, consisting of keys &amp; values, by value. For example, I have a hash of words and respective frequencies, that I want to order by frequency.  There is a SortedList which is good for a single value (say frequency), that I want to map it back to the word.   orders by key, not value. Some resort to a , but is there a cleaner way? Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.   $(a,b)=1$ means $a$ and $b$ have no prime factors in common  $ab$ is simply the product of factors of $a$ and factors of $b$.  Let's say $k\mid a+b$ where $k$ is some factor of $a$.  Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.  So $a(k-l)=b, \  a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].  But $(a,b)=1$ so $a$ cannot divide $b$.  We have a similar argument for $b$.  So $a+b$ is not divisible by any factors of $ab$.  Therefore, $(a+b, ab)=1$.  Would this be correct? Am I missing anything? The reason that I ask is because this is happening to LiDAR, and I am inclined to use lidar because it is easier to read.