Convergence of $\sum\limits_{n=2}^{\infty}\frac{(n^3+1)^\frac13-n}{\log n}$ To show that $$\sum\limits_{n=2}^{\infty}\frac{(n^3+1)^\frac13-n}{\log n}$$ is convergent My attempt: Expanding the first term using binomial series and simplifying, I tried comparison test with $b_{n}=\frac{1}{n^2}$  $$\lim_{n \to \infty}\left(\frac{(n^3+1)^\frac13-n}{\log n}\right)n^2 $$ $$=\lim_{n \to \infty}\frac{n\left(\frac13\frac{1 }{n^{3}}-\frac19\frac{1 }{n^{6}}+\dots\right)n^2}{\log n}$$ I could simplify this using binomial theorem but I don't know how to deal with the $\log$ term in the denominator
Show that the series $\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$ converges Show that the series $$\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$$ converges.   I showed it using Abel's theorem and limit comparison test. Any other simpler method?
I'm studying for actuarial exams, but I always pick up mathematics books because I like to challenge myself and try to learn new branches.  Recently I've bought Topology by D. Kahn and am finding it difficult.  Here is a problem that I think I'm am answering sufficiently but any help would be great if I am off.  If $d$ is a metric on a set $S$, show that  $$d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$$ is a metric on $S$.  The conditions for being a metric are $d(X,Y)\ge{0}, d(X,Y)=0$ iff $X=Y$, $d(X,Y)=d(Y,X)$, and $d(X,Y)\le{d(X,Z)+d(Z,Y)}$.  Thus, we simply go axiom by axiom.  1)  Since both $d(x,y)\ge{0}$ and $1+d(x,y)\ge{0},$ it is clear that $d_1(x,y)\ge{0}$.  (Is this a sufficient analysis?) 2) $d_1(x,x)=\frac{d(x,x)}{1+d(x,x)}=\frac{0}{1+0}=0$. 3) $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}=\frac{d(y,x)}{1+d(y,x)}=d_1(y,x).$ 4) $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}\le{\frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}}=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}\lt\frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}=d_1(x,z)+d_1(z,y).$  However, #4 is strictly less, not less than or equal to, according to my analysis, so where did I go wrong? Why do DNGs only show in low resolution on Google Photos? Through my NAS at home I store and backup all my photos on Google Drive. These photos also show up on Google Photos. I use Lightroom and convert all photos to DNG when copying to my NAS.  When I look at the photos in Google Photos the image is rather small. Clicking the (i) info button tells me the image has a resolution of 1024x680, file size 9.8 MB and a MP of 0.7. This is probably just a preview or thumbnail of some sort, but how do I view the image in larger resolutions in Google Photos? On the  it is mentioned that it is not a Cayley graph.    How it this proved? Honestly I don't even know how to start this. The only criteria I can think of is that all vertices must have the same degree. Also the degree being odd should imply that one generator of the group has order 2. But then how do I proceed?  Edit: As already mentioned in the comments, the link did not answer my question. A vessel is filled with liquid,3 parts of which are water and 5 parts syrup.How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?  What is the best possible way to solve this question? Can I get a certificate from a root CA that I can then use to sign my own web server certificates? I would, if possible, use a signed certificate as an intermediate to sign other certs.  I know that I would have to configure my systems in a certain way with "my" intermediate certificate in order to supply information about the chain of trust to my clients.   Is this possible? Are root CAs willing to sign a certificate like this? Is it expensive?  BACKGROUND  I'm familiar with the basics of SSL as it pertains to securing web traffic over HTTP. I also have a basic understanding of the way the chain of trust works, in that web traffic is secured "by default" if you encrypt with a certificate that has a valid chain all the way back to a root CA, as determined by the browser/OS vendor.  I am also aware that many of the root CAs have begun signing certificates for end users (like me) with intermediate certificates. That may require a bit more setup on my end, but otherwise, those certificates will work fine. I guess this has to do with protecting their all-valuable private key for the CA and the disaster that it would be if i were ever compromised.  EXAMPLES        Now, we are definitely not the size of any of those organizations, but they seem to be doing something like this. It would definitely make the management of these certificates a lot more palatable, especially considering one way we are expanding the reach of our e-commerce platform. Custom 500 Error page: Is there a way to theme a fatal error? In /core/includes/errors.inc, when there is a fatal error it outputs a response with the message "The website encountered an unexpected error. Please try again later." and a back-trace, if error_level is set to verbose, and without the back-trace otherwise.  It seems like at this point, there's no way, with an event subscriber to capture the response and alter it.  Is that correct?  Or is there a way to wrap the output from line 268 of errors.inc in Drupal 8.4 with markup?    Note, there's another , but that doesn't offer a complete solution.  Also, I looked into .htaccess and Apache ErrorDocument directive, but that . Once Apache passes off to PHP, it can no longer use that directive.  That's for real server errors. How can I make the Logitech Wingman Formula Force steering wheel work in Vista and 64 bit Windows 7? Most of the time, it should be really easy to determine who is my ally and who is not. The people who try to kill me are most likely my foes, whereas the people who entered the dungeon with me are probably my allies.   But sometimes it’s not that simple. What if one of my party members got replaced by an evil doppelganger, who plans to kill me in the near future? What if one of my allies infiltrated the court of the evil king and I accidentally have to fight her, while she is (unbeknownst to me) under the effect of a  spell? What if a character fell in love with the BBEG and just wants to save him from himself, while the BBEG hates the character and actively tries to murder him?  Most spells and abilities that affect others specify the targets as “creatures” or “willing creatures”. Some explicitly affect allies only. That’s why I want to know, how to determine who’s allied with whom. Do I have to think that someone is my ally for someone to count as my ally? Does someone else have to believe to be my ally in order to count? Do both of us have to agree on being allies? Is there an objective observer who determines who’s allied? Can Person A be the ally of Person B, while Person B is not the ally of Person A?  There are many examples for spells/abilities that care about allies, among others:   the Exalted Champion feature from the  (SCAG, p. 132-133), that grants allies within 30 feet advantage on Wisdom saving throws. the Inspiring Surge feature from the  (SCAG, p. 128), that allows one or two allies within 60 feet to make a melee weapon attack as a reaction.   With features like this, you could accidentally affect someone who’s not actually your ally. You could fail to affect someone who is actually your ally, but you don’t know it (yet). Or you could affect someone who doesn’t want to be affected (because they don’t want to be allied with you). This all depends on the interpretation of the word “ally”. I’m not sure if "ally" is a defined game term (but I don’t think so), or if I have to rely on the dictionary definition. That’s why I want to ask:  How do I determine who’s allied with whom? How do I deal with a bored or distracted player bringing down the group? My group have been playing various D&amp;D style games for a few years,  I have one player however that often will drift off and start playing on his phone  and several times has actually fallen asleep (with his character having to be "nudged" along by his fellow players).  He insists he's "just tired" and "isn't bored" — and besides, he keeps insisting we all get together for D&amp;D and wants to play.  Unfortunately we're all pretty busy, so if I'm to get the group together at all it's usually late on a Sunday. After talking to the player he insists it's not the game, or the story, he just happens to be tired. But, it's the same excuse every week, and the odd times we've played at different times or situations, the falling asleep has happened.  How do I cope with:   The morale destroying (for me) idea of your players falling asleep through a campaign and story you've put lots of work into. The general tone of the room when this player loses interest. (Occasionally a couple of others zone out into Facebook when they're not "active" because the players have split up / gone unconscious).   Given that it's an "all or nothing" situation with the group (there are only 3 players and they won't play unless all 3 are around), how do I best deal with it? how to calculate $\int_C \frac{2xy^2dx-2yx^2dy}{x^2+y^2}$ using green theorm or directly Calculate $$\int_C \frac{2xy^2dx-2yx^2dy}{x^2+y^2},$$ where $C$ is the ellipse $3x^2 +5y^2 = 1$ taken in the positive direction.   I tried to calculate the integral using green theorm.  now i need to build enclosier that doesn't enclose $(0,0)$ i am having hard time guessing what to build.  a circle and ellipse might be perfect but then the domain is not easy to write. can i have hint please ? author not appearing in \maketitle in amsart I am new to the amsart document class. For some reasons \author (etc..) is not showing after I use \maketitle. This is my tex:  \documentclass[12pt,reqno,a4paper]{amsart}  \usepackage{amsmath} \usepackage{amssymb} \usepackage{amsfonts} \usepackage{mathrsfs}  \usepackage{amsthm}  \usepackage{latexsym}  \usepackage{pictexwd} \usepackage{pdftricks} \usepackage{natbib} \usepackage{setspace} \usepackage{color} \usepackage[affil-it]{authblk}  \begin{document}  \title{Temptation, Asymmetric information and Committment}  \author{Mark}  \affil{University of Leicester}  \date{February 2014}  \maketitle  \begin{abstract} To be written \end{abstract}  \section{introduction}  ...  \end{document}   and this is the output:   can anybody help me out? Short brandable domain vs. long keyword domain I need a counsel about my websites' domain.  Let's say I have just bought two domains: xyz.com and xyzphotography.com.  As we can see, the first one is shorter but way less meaningful than the second one. The second one contains a photography keyword but it's longer.  I have only one website which is my photography portfolio. Can you advise me how to set those domains to make it all as SEO-friendly as possible? Which one should be the main one, does the rest should work on their own or have some kind of SEO-friendly redirect to the main domain? Show that $e^{-\beta} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-\beta^2 / 4u} du$. Show that $e^{-\beta} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-\beta^2 / 4u} du$.  I'm not really sure of how I should proceed to show this, and it's pretty un-intuitive as well.  I've managed to manipulate the RHS into  $$\frac{e^{4\beta}}{\sqrt{\pi}} \int_0^\infty \frac{1}{\sqrt{u}} e^{-(\beta+2u)^2/4u} du$$ which now vaguely resembles a Gaussian. However, substituting $v = \beta + 2u$ in in order to to get something like $e^{-x^2}$ doesn't particularly help. I got $$ \frac{e^{4\beta}}{\sqrt{2\pi}} \int_0^\infty \frac{1}{\sqrt{v - \beta}} e^{-v^2/2(v-\beta)} dv. $$  I'm really unsure of how I should proceed; any help would be appreciated. I'm attempting to get a better understanding of cryptography and have run into some questions about bcrypt.   Bcrypt relies on the blowfish cipher, which is a reversible method of encryption. But bcrypt is not reversible. How is that possible? Or am I mistaken? What does bcrypt do on top of blowfish to make it irreversible? Why does bcrypt use blowfish instead of twofish or threefish? Hello can someone give me short tutorial how to:   Remove Windows 7 from my dual boot system (Ubuntu/Windows7) Let me use free space as new partition for data (so I don't need to extend Ubuntu itself here) Remove windows 7 option from boot screen   Update:  Thanks to Lekensteyn answer I successfully deleted windows. Now I created new partition (formatted to ext4), but I can not use it. I cannot copy files there, only as root.  Also when booting, Ubuntu complains it didn't find /windows partition (of course it didn't, I removed it). I feel like I should update some tables or what, any ideas? How can I get to Iceland on a cargo ship? Planning a low budget and environment friendly trip to Iceland from Helsinki, Finland next summer for me and my friend. Haven't been able to find relevant information on possibility for cargo travel (from where - eg. Norway, Denmark? - timetables, prices etc). Expected departure from Helsinki around June 18th, return at latest on July 7th 2015. I'm very thankful for any related advice (and first time on stackexchange, so apologies if I didn't formulate my question in a best possible way) Novel from 50s or 60s - Good vs. Evil I'm not sure I recall enough of this story for a good identification, and the below is definitely out of story order, but...  The POV character goes to his home town and finds it different. He later finds that he's trapped in the town; there are two young people who turn out later on to be ... avatars ... of Ormazd and Ahriman, and the town is a battleground; what he is seeing of the town is Ahriman's effect on the town, and he turns out to have some limited ability to see the original town, which is more-or-less Ormazd's version of reality. In the end, Ahriman's avatar is destroyed, and the town restored.  I read this in the '80s or earlier; it was part of my father's collection, and the book was showing its age; I think it more likely to be the '60s rather than the '50s, and was probably by a prominent name in SF, but not one of the Masters. $$I(\alpha)=\int_0^1 \frac{x^\alpha-1}{\ln x}dx.$$  As the title says, if someone could solve this without using the differentiation under the integral sign technique, I would be very grateful. The situation is my  has been enchanted with .  The creature has      when it dies, return it to his owner’s hand   The enchantment says     When enchanted creature dies, return that card to the battlefield under your control with a +1/+1 counter on it.   Who gets the creature? Me or my rival?  Does the creature ability trigger before or after enchantment? Number of pairs of strings satisfying the given condition We are provided with a string A (may contain repeated characters as well ). One needs to make all possible permutations of A , and find out how many pairs of strings chosen from these permutations are not similar.  Two strings are similar if we can make both of them exactly same by carrying out at most 1 swap in each of them .  Example :- if we are provided with string abcd  , then for permutation abcd , the permutations which are not similar to it are bcda , bdac , cadb , cdba , dabc , dcab . Similar is the case for other permuations as well. So overall number of desired pairs are 24*6=144 .  I tried solving it for each possible pair by approach mentioned here     But as the permutations can be large enough , I need a better approach of solving this.