$\frac1{\sqrt{1}} + \frac1{\sqrt{2}}+\frac1{\sqrt{3}}+...+\frac1{\sqrt{x}}\ge {\sqrt{x}}$  I proved the basic case: and realize it is equal to 1, but I have absolutely no idea how to create prove the left and right side using the induction hypothesis.  Please help :) I've solved equality's before using induction, but this is the first time I've done an inequality, and although I'm sure the process is fairly similar, I think the square root is messing me up.
I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing.  $$\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$$  Perhaps someone can help me out I don't understand how to move forward from here: $$\sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n+1}}\ge \sqrt{n+1}}$$ proof and explanation would greatly be appreciated :) Thanks :)  EDIT sorry meant GE not = fixed :)
Spurious space above align environment at top of page/minipage It seems that the align environment adds additional vertical space if it starts at the top of a page or minipage.  Below are two pages, and two minipages (on Page 1), where the first page/minipage has text followed by an align* environment, and the second page/minipage starts with an align* environment, and note that the equations are still aligned.  Attempted Solution: The closest related question I found was:    I attempted to use the  at the above by adding \setlength\abovedisplayskip{0pt} (after the preamble as per ).  Although, this did move the equations up vertically, they were still vertically aligned:  And since the problem occurs at top of a page as well, I don't think this is related to  Question:  How can I adjust the align environment to automagically not have additional space at the top of a page or minipage? Is there a reason for this behavior, and perhaps should not be changed?  Code: \documentclass{article} \usepackage[showframe,paperwidth=3.5in,paperheight=2.2in,margin=0.5in]{geometry} \usepackage{amsmath}  \begin{document}%\setlength\abovedisplayskip{0pt} Top of Page 1 \begin{align*}% Text above at top of page  e &amp;= m c^2 \end{align*}  \hrule \begin{minipage}[t]{0.45\linewidth}     Top of minipage 1     \begin{align*}% Text above this minipage      e &amp;= m c^2     \end{align*} \end{minipage} \begin{minipage}[t]{0.45\linewidth}     \begin{align*}% No text above in this minipage      e &amp;= m c^2     \end{align*} \end{minipage}  \newpage \begin{align*}% No text above at top of page  e &amp;= m c^2 \end{align*} \end{document} Given a simple graph and its complement, prove that either of them is always connected. I was tasked to prove that when given 2 graphs $G$ and $\bar{G}$ (complement), at least one of them is a always a connected graph.  Well, I always post my attempt at solution, but here I'm totally stuck. I tried to do raw algebraic manipulations with # of components, circuit ranks, etc, but to no avail. So I really hope someone could give me a hint on how to approach this problem. What is the limit of this sequence? How does it relate to the exponential function? Find $$\lim_{n\to\infty} \left(1+\frac{x^2}{n^2}\right)^n$$   I particular, I am hoping to find the above to be $1+x^kg(x/n)$ where $g(\cdot)$ is a function with uniformly bounded derivatives.  Edit: I'm trying to prove a result that would be true if  $(1+x^2/n^2)^n -1 = x^kO(x/n).$ Title does not specify "On Hold" Testing the app, I noticed the title does not specify if a question is "On Hold" ... I haven't checked specifically, but would assume it doesn't specify as "Duplicate" either ... Or am I missing it somewhere? Is it not in the title but located/displayed some other way? If it is, this could be very confusing to users. If it isn't there, it's definitely confusing to users :-)  Using: Version 1.0.1.125 on iPad3 with iOS8 $(U\cap W)^0 = U^0 + W^0$ Let $V$ be a finite dimensional vector space and $U^0$ and $W^0$ denote the annihilators of subspaces $U$ and $W$, respectively.  Then I need to show that $(U\cap W)^0 = U^0 + W^0$.  I can show that $U^0 + W^0 \subseteq (U\cap W)^0$ like so: Let $f\in U^0$ and $g\in W^0$.  Then, because $U^0, W^0 \subseteq (U\cap W)^0$, we see that if $f\in U^0$ and $g\in W^0$, then for any $v\in U\cap W$ and $a,b \in \Bbb F$, $(af+bg)(v) = af(v) + bg(v) = 0$.  Thus $af+bg\in (U\cap W)^0$.  But because $f$ and $g$ were arbitrary, this implies that any linear combination of elements of $U^0$ and $W^0$ is an element of $(U\cap W)^0$.  I.e. $U^0+W^0 \subseteq (U\cap W)^0$.  But I haven't had any luck proving the other direction.  Obviously I should start with $f\in (U\cap W)^0$.  But then it looks like I need to somehow break that function into two to show its in $U^0 + W^0$.  That's throwing me off. If asking a question or giving instruction towards a group of objects where the number is unknown (and possibly could be one) what is the correct way to phrase the sentence? Which of the following sentences are more correct:     Has there been any studies regarding this and if so what has been the   conclusion?   OR     Has there been any studies regarding this and if so what have been the   conclusions? All of my Firefox extensions were spontaneously and simultaneously disabled (converted to Legacy status) a few moments ago.  How can I re-enable them? Firefox 66.0.3 (64-bit) I have a blender file where a sphere is orbiting a force field. I want it to be more red when it's closer to the center, and more blue when it's further. I already have the math figured out, I just need to know how to get x/y coordinates in nodes.  Here's the file: What will be the one's digit of the remainder in: $\left|5555^{2222} + 2222^{5555}\right|\div 7=?$ Limit of a decreasing sequence of measurable sets. Let $(X,\mathcal{A})$ be a measurable space, with measure $\mu$. Let $\{E_n\}_{n \in \mathbb{N}} \subseteq \mathcal{A}$ be a sequence of measurable sets, with $E_{n+1} \subseteq E_n, \ \forall n \in \mathbb{N}$, that is a decreasing sequence, and $\mu(E_n)=+ \infty, \ \forall n \in \mathbb{N}$. Let $E=\bigcap_{n \in \mathbb{N}} E_n$ the limit set of the sequence, $E$ is measurable for definition. Is true that $\mu(E)=+\infty$? If not in all cases, is true with $X=\mathbb{R}^{N}$, $\mathcal{A}$ the collection of Lebesgue-measurable sets, and $\mu$ the Lebesgue measure on $\mathbb{R}^N$? This is an unofficial list/changelog of new features and various changes to Stack Overflow and the Stack Exchange network.  It is maintained by the community, while a Stack Exchange employee changes the accepted answer to ensure that the latest changes remain on top (given default user settings).  To see the list in reverse chronological order (latest entry on top), .  For those adding new features here: please post only feature changes that affect the whole network and have some global impact. i.e. bug fixes don't belong here, nor do mobile app updates. The axiom of choice asserts that for any set $X$ there exists a function $f : (2^X − \{\emptyset\}) \rightarrow X$ such that for any nonempty $A \subseteq X$, $f(A) \in A$. Show that this is equivalent to the assertion that every relation includes a function with the same domain. I am given this problem in my assignment. My approach is following. I have a set $X$. Then it has a choice function $g : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$. Suppose $X$ has a relation $R \subseteq X \times X$. $dom(R) \subseteq X$. Then the required function $f : dom(R) \rightarrow X$ is $$f(x) = g(\{y\ |\ (x, y) \in R\})$$ Conversely, suppose $X \supseteq Y$ is a non-empty subset. Let $$R_Y = \{(x, y)\ |\ y \in Y, x \in Y, (p, q) \in R_Y \Rightarrow p = x\}$$ Then there is a function $f_Y : dom(R_Y) \rightarrow X$. But $dom(R_Y)$ is singleton from the definition of $R_Y$. Then $im(f_Y)$ is also singleton. We define our choice function as $f = \{(Y, y)\ |\ y \in im(f_Y)\}$ Are my arguments correct?  Edit.1: what I tried to do for converse is, for any subset $Y$, I construct a relation $Y_1 \times Y$ where $Y \supseteq Y_1 = \{y\}$. Then there is a function $f_Y : Y_1 \rightarrow Y$. Then if $f$ is my choice function I define $f(Y) = f_Y(y)$. May be I should define $R_Y$ as $$R_Y = \{(x, y)\ |\ y \in Y; x \in Y; (p, q), (p', q) \in R_Y \Rightarrow p = p'\}$$ I understand there may be multiple $R_y$'s. But can't I define a singleton set as $\{s\in S\ |\ a, b \in S \Rightarrow a = b\}$? This can be $\{s\}$ $\forall s \in S$. But isn't it a valid specification?  Edit.2: This is the final solution with the help from @Asaf.  If we assume axiom of choice then for any $X$ there is a choice function $f : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$ such that $f(A) \in A$. Suppose $R$ is a relation between $P$ and $Q$. Then we have a choice function $f_Q$ for $Q$. Let $f_R(p) = f_Q(\{q\ |\ pRq\})$. The function $f_R : P \rightarrow Q$ and $dom(f_R) = dom(R)$. So this is our required function.  Conversely, if every relation includes a function with same domain. Let $R$ be a relation between $\mathcal{P}(X) \setminus \left\{\emptyset\right\}$ and $X$ such that $aRb \Rightarrow b \in a$. Now $R$ includes a function $f : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$. Then $xRf(x) \Rightarrow f(x) \in x$. So $f$ is the choice function for $X$. How can I search within a manpage? Where can I ask questions regarding Firefox? I can't seem to find info on installing a Swedish language package on my machine. The language is not listed  and the hits on Google I got led nowhere (or, rather, to the said page).  Where can I ask such a question (within the realm of Stack Exchange)? In an electromagnetic theory class, my professor introduced the concept of "action at a distance in physics". He said that:     If two charges are at some very large distance, and if any one of the charge moves, then the force associated with the charges changes instantaneously.   But according to Einstein, no information can travel faster than the speed of light.   So photons (the information carriers in electromagnetic force) cannot instantaneously deliver information.   So that we associate a field with the two charges and if any charge moves, there is a deformation in that field and this deformation travels with the speed of light and conveys the information.   If the field deformation information cannot travel more than the speed of light, how does the force instantaneously change at very large distances? Batch Processing Quick Query I have like 50000 records that meet my condition for batch processing. My batch process begins at say 10:01 AM and ends at 10:10 AM. I set my batch size to be 200.  Can someone tell me what would happen  a) If an user creates a record at 10:03 AM that meet my condition...will this record be also be processed or not ?  b) Will the records which are getting processed during Batch 1 are technically locked till that particular batch is processed completely ?  c) What will be the behaviour if a record in Batch N does not meet my condition anymore but that particular batch has already been started to process ?  d) Is it possible to create a parallel processing of sorts such that Salesforce (instead of running one batch at a time) it will start processing parallely with multiple batches ?  I searched online and I could find manual/instructions with respect to how to run batch apex but I am not able to get clarity with respect to these specific cases. The discrimination between short and long option is only `-`? On Unix system, options begin with - is short option, while options begin with -- is long option. Is my understanding right? How to interpret a quantum circuit as a matrix? Verb form of "must needs" Then touching the shoulder of a townsman who stood near to him, he addressed him in a formal and courteous manner:      "I pray you, good Sir," said he, "who is this woman? -- and wherefore is she here set up to public shame?"      "You must needs be a stranger in this region, friend," answered the townsman, looking curiously at the questioner and his savage companion, "else you would surely have heard of Mistress Hester Prynne and her evil doings. She hath raised a great scandal, I promise you, in godly Master Dimmesdale's church. "   As far as I know we always use first form of the verb after "must". For instance: "You must go". We never say "You must goes". Also, we never use -s/es with a verb that comes with "You". What are the circumstances in which writer has used "needs" after "You must" in "You must needs to be a stranger in this region."? What is difference between a ring and a field?