How to show that $ \sum_{k=1}^{2^n}\frac{1}{k}\geq 1+\frac{n}{2}$? So in order to prove that the Harmonic series diverge, I want to show that: $$ \sum\limits_{k=1}^{2^n}\dfrac{1}{k}\geq 1+\dfrac{n}{2}$$ It is clear that if we expand the sum that this inequality holds true, $$n=3,\ \sum\limits_{k=1}^{2^3=8}\dfrac{1}{k}=1+\frac{1}{2}+\underbrace{\frac{1}{3}+\frac{1}{4}}\limits_{\geq \frac{1}{2}}+\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}\limits_{\geq \frac{1}{2}}$$ but how can i show that using induction?  Any tips would be helpful, and thanks.
Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$ Prove by induction the summation of $\frac1{2^n}$ is greater than or equal to $1+\frac{n}2$.  We start with $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$$ for all positive integers.   I have resolved that the following attempt to prove this inequality is false, but I will leave it here to show you my progress. In my proof, I need to define P(n), work out the base case for n=1, and then follow through with the induction step. Strong mathematical induction may be used.  This is equivalent to $$\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n}2\;.$$  Let $P(n)$ be summation shown above.   Base case for $n=1$, the first positive integer,   $$\sum_{k=0}^1\frac1{2^k}=\frac1{2^0}+\frac1{2^1}=1+\frac12=\frac32\ge 1+\frac12=\frac32\;,$$  so base case is true.  Induction step: Assume $P(n)$ is true and implies $P(n+1)$. Thus   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge\frac1{2^{n+1}}+\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n+1}2\;.$$  This can be written as   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge \frac1{2^{n+1}}+1+\frac{n}2\ge 1+\frac{n+1}2\;.$$  I work the math out but I get stuck contradicting my statement. Please show your steps hereafter so I can correct my mistakes.
Would the One Ring even work for anyone but Sauron? What does the comma operator , do? Kaplansky's theorem of infinitely many right inverses in monoids? There's a theorem of Kaplansky that states that if an element $u$ of a ring has more than one right inverse, then it in fact has infinitely many. I could prove this by assuming $v$ is a right inverse, and then showing that the elements $v+(1-vu)u^n$ are right inverses for all $n$ and distinct.  To see they're distinct, I suppose $v+(1-vu)u^n=v+(1-vu)u^m$ for distinct $n$ and $m$. I suppose $n&gt;m$. Since $u$ is cancellable on the right, this implies $(1-vu)u^{n-m}=1-vu$. Then $(1-vu)u^{n-m-1}u+vu=((1-vu)u^{n-m-1}+v)u=1$, so $u$ has a left inverse, but then $u$ would be a unit, and hence have only one right inverse.  Does the same theorem hold in monoids, or is there some counterexample? Get exit status of process that's piped to another I have two processes foo and bar, connected with a pipe:  $ foo | bar   bar always exits 0; I'm interested in the exit code of foo. Is there any way to get at it? I've got a NI Traktor Audio 2 sound card I'd like to get working with Mixxx, but I don't seem to have any success so far. The card seems to be supported in Ubuntu out of the box, and it seems to work well with ALSA, but I cannot select it in Mixxx and I'm not sure whether is something I can do to make it see it (e.g. through).  So for testing purposes, on the command line I can execute aplay to check each channel:  aplay -D plughw:2,0,0 test.wav   and  aplay -D plughw:2,0,1 test.wav   That works well: I can hear the test sound on each channel (A and B)  The relevant output for this card using aplay -D is:  hw:CARD=TraktorAudio2,DEV=0      Traktor Audio 2, Traktor Audio 2      Direct hardware device without any conversions plughw:CARD=TraktorAudio2,DEV=0      Traktor Audio 2, Traktor Audio 2      Hardware device with all software conversions   From those 2 items on the output, plughw is the one that works, whereas hw doesn't.  The problem is that Mixxx only lets me select the hw card, which does not seem to work. I'd like to be able to select the plughw card, which is the one that's working.  I've tried suspending Pulse Audio as suggested by , but that did not seem to work. I also played around with an .asoundrc file similar to the one  (which was for an older model of this sound card), but also without success.  Any ideas on how to make Mixxx show the right card (and channel, while we are at it)?  Thanks! My Ubuntu laptop which was working fine until recently. My desktop disappeared and all the desktop icons are gone. All I have access to is the terminal. In an attempt to fix the problem, I installed GNOME but it made the computer give the error:     Warning! Failed to connect to lvmetad. No such file or directory Falling back to internal scanning.    I can press Ctrl+Alt+F1 to start tty1 but I can't get to a graphical user interface. What did I do wrong and how do I fix it?.  The operating system is ubuntu Linux 'Possible duplicate' comment not appearing for questions with deleted comments containing the link I posted a comment to a duplicate question on . But I decided to delete it. Afterwards I voted as duplicate and thought a comment would be generated like     possible duplicate of ...   But it didn't. My guess is that the duplicate-comment won't be generated if a comment exists with that link, even if deleted. Gibbs free energy intuition What is ? As my book explains:     Gibbs energy is the energy of a system available for work.   So, what does it want to tell? Why is it free? Energy means ability to do work. What is so special about this energy? Can anyone simply explain?   I just want a math-free intuition. I was reading a  where the author mentions: "The coefficient of variation is primarily a descriptive statistic, but it is amenable to statistical inferences such as null hypothesis testing or confidence intervals. Standard procedures are often very dependent of the normality assumption and current work is exploring alternative procedures which are less dependent on this normality assumption."  The paper is from 2010. So my question is: has there been any recent "advancements" in terms of a coefficient of variation statistic that is not dependent upon the normality assumption? And, is this what one may call a non-parametric CV? I tried for a while to classifiy the singularities of $\frac{1}{z}-\frac{1}{\sin z}$ at the origin, but I am stucked. Is there someone who is able to help me at this point? how to get the string-lengths stored in array (C) The set of convergent points of a sequence of continuous functions is Borel Suppose that $f_n$ are continuous.  Prove that $E=\{x\in \mathbb{R}:f_n(x)\text{ is convergent}\}$ is Borel.   My first instinct is to take balls around $\lim_{n\to\infty}f_n(x)$, for each $x$ in $E$, and then show that there has to be some ball small enough so that everything inside that ball is a member of $E$. But this would imply not only that $E$ is Borel, but moreover that it is open, so I don't think that can be right.  Can someone show me how to do this? By definition a matrix representing a Lorentz transformation is orthogonal, so that its inverse is equal to its transpose.   Consider a pure boost in the t-x plane; $$\Lambda_x=\begin{pmatrix} \cosh(\gamma) &amp;&amp; \sinh(\gamma) &amp;&amp; 0 &amp;&amp; 0\\ \sinh(\gamma) &amp;&amp; \cosh(\gamma) &amp;&amp; 0 &amp;&amp; 0\\ 0 &amp;&amp; 0 &amp;&amp; 1 &amp;&amp; 0\\ 0&amp;&amp;0&amp;&amp;0&amp;&amp;1 \end{pmatrix}.$$ $\Lambda_x$ has inverse  $$\Lambda_x^{-1}=\begin{pmatrix} \cosh(\gamma) &amp;&amp; -\sinh(\gamma) &amp;&amp; 0&amp;&amp;0\\ -\sinh(\gamma) &amp;&amp; \cosh(\gamma) &amp;&amp; 0&amp;&amp;0\\ 0 &amp;&amp; 0 &amp;&amp; 1 &amp;&amp;0\\ 0&amp;&amp;0&amp;&amp;0&amp;&amp;1 \end{pmatrix}$$ but tranpose  $$\Lambda_x^T=\begin{pmatrix} \cosh(\gamma) &amp;&amp; \sinh(\gamma) &amp;&amp; 0&amp;&amp;0\\ \sinh(\gamma) &amp;&amp; \cosh(\gamma) &amp;&amp; 0&amp;&amp;0\\ 0 &amp;&amp; 0 &amp;&amp; 1&amp;&amp;0\\ 0&amp;&amp;0&amp;&amp;0&amp;&amp;1 \end{pmatrix}.$$ These are not equal. Where have I gone wrong? Identify which OrgWideEmailAddress records are valid for current user I'd like to create a popup list on one of my Visualforce pages, for a user to select from a list of OrgWideEmailAddress records.  My goal is to limit the list to only those records that they're allowed to use.  I'm going to save the selection for use in later automatic emails.  Based on what I've read, I won't be able to send emails to certain OrgWideEmailAddresses:   The address hasn't been verified yet.  I don't see any fields in the schema that appear to show verification state. The address is only available to profiles that the user isn't a member of.  I don't see where the profile list is stored for each record.   Some of my problem could just be due to not understanding the standard objects so well, and perhaps there's a way to get at this through UserInfo, or by making my query on the User object which may have a child relationship to OrgWideEmailAddress since it's listed in the schema viewer of the Force.com IDE.  Here's the test code I'm working with:  for (OrgWideEmailAddress addr : [SELECT DisplayName, Address          FROM OrgWideEmailAddress ORDER BY DisplayName]) {     System.debug( addr.DisplayName + ' &lt;' + addr.Address + '&gt;'); }   I tried creating a new Messaging.SingleEmailMessage and then calling setOrgWideEmailAddressId() on it with each OrgWideEmailAddress Id, in hopes that invalid Ids would throw an exception, but to no avail.  My goal is to modify the test code to provide just the subset of records that the current user will be able to use. Simple examples of uncorrelated but not independent $X$ and $Y$ There is a lemma that says if a group $G$ has no proper nontrivial subgroups, then $G$ is cyclic. And here is the proof of the lemma:     Suppose $G$ has no proper nontrivial subgroups. Take an element $a$ in $G$ for which $a$ is not  equal to $e$. Consider the cyclic subgroup $\langle a \rangle$. This subgroup contains at least $e$ and $a$, so it is not trivial. But $G$ has no proper subgroups, so it must be that $\langle a \rangle = G$. Thus $G$ is cyclic, by definition of a cyclic group.   But here i do not understand the following: Why must $\langle a \rangle$ be a subgroup of $G$? For every single element $a$ in $G$, if $\langle a \rangle$ is a subgroup of $G$, then every group should have at least as many subgroups as the number of its elements. I would appreciate any help. Thanks Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers? $$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$  I tried AM-GM, CS inequality but all failed. Markdown for ordered lists, starting with zero? I see that you can create an ordered list using Markdown notation like the following.  1. line 1 2. line 2 3. line 3 ... N. line N   The Markdown content supports starting each line with any arbitrary number and reorders the list starting with 1. I'd like to start an ordered list with zero. My use case for zero would be in describing capture groups from regular expressions where group zero always returns the entire string. So I'd like to display something like this:  # Groups 0. Entire string 1. The first capture group you were looking for 2. The second capture group 3. The third group with the string I heart Kittens   Markdown instead reorders the group and forces the ordered list to start at 1:    Trying to force the issue with inline HTML: &lt;ol start="0"&gt; or &lt;ol&gt;&lt;li value="0"&gt;entire string&lt;/li&gt;&lt;/ol&gt; doesn't work either. Any ideas on how I can create ordered list starting with zero?  If it's not possible to do so, then this is a feature-request. What is the plural of the abbreviation of "multiplicity automaton", "MA" or "MAs"? The "multiplicity automaton (MA)" is a model in compute science and its plural is "multiplicity automata". Should the plural of the abbreviation be MA or MAs?      is not a duplicate as it is focused on cases like "ATMs" where the expanded form "automated teller machines" has a regular plural ending in "s." Unlike "machines," the word "automata" has an irregular plural that does not end in "s." It's unclear from the linked answers how abbreviations that end in words like this should be pluralized. A high-ranking  mentions "VIPs," but a high-ranking  says     VIPs' is a strange one. It could be argued that the expanded acronym,   when pluralized, should be 'Very Important People' rather than 'Very   Important Persons'. Using this argument, 'VIP' could be used as a   singular or plural noun. "The VIP has arrived." "The VIP have   arrived." I'm not sure anyone else would agree with my logic, though. Odds of rolling doubles or triples given N 6-sided dice Over the holidays I played a dice game where each player had 3 to 7 d6 to roll each turn.  The game gave certain advantages to doubles and triples.  I wanted to know the odds of rolling doubles or triples given N dice to understand the importance of "upgrading" to more dice.  My question is really about how to reason about such problems:  It's easy to see that the odds of rolling doubles with 2 d6 are 1/6 (first die can be anything, and 1/6 of the time the second will match).  It's also easy to see that with 7 dice the odds are 1.0 (6 dice could be 1..6, but the 7th must match one of them).  The middle cases are fuzzier to me.  I ended up writing a program to enumerate all of the possibilities (even 6^7 is only 279,936 cases to evaluate) because that was the only way I felt I could verify any closed formula I came up with.  This also enabled me to distinguish cases of multiple doubles (or triples).  I would like to know how to derive closed forms and how to validate such solutions without brute-force simulation.