I've seen multiple ways on how to solve it online (and most likely the majority are wrong), but I don't know how to solve this fully so that I could get the full grade during my exam.  The question is as follows:     Prove that if $\sum a_n$ is convergent with $a_n &gt; 0$ for all $n$,   then $\sum a^2_n$ is also convergent.   I understand that the starting point to prove this is that $\lim a_n =0$, but after that I really don't know what I'm supposed to say.
Let $(a_n)$ be a sequence of positive terms and suppose that $\sum_\limits{n=0}^{\infty} a_n$ converges. Show that $\sum_\limits{n=0}^{\infty} a_n^2$ converges.   This is in the section on the Comparison Test so that must be what I'm supposed to use.  But I don't see how.  $(a_n)^2$ might be smaller or larger than $a_n$ depending on $a_n$.  And I can't use the Comparison Test with some other series because there's no info here about how fast $\sum a_n$ converges.  Hmm.  Any hints?
Force Messages to use SMS for specific contact rather than iMessage? Essentially, I would like to set my phone so that all my outgoing messages to a specific friend, with whom I currently text back and forth over iMessage, are sent out over SMS, never iMessage. Additionally, I would like to block any incoming iMessage texts of his and (hopefully) they'll fail and be resent via SMS.  Let me restate that: I would like to force SMS for both parties, going both ways, for a specific individual contact.   Is this possible? Blocking any usage of iMessage between myself and this other individual, so they must either send and receive any texts to or from me via SMS (or the texts fail/never arrive). First, this question is motivated by the imprecise question: Is there a sensible notion of parity (evenness and oddness) for real numbers?   Here are some properties a notion of parity for $\mathbb{R}$ should have:    It should be an equivalence relation on $\mathbb{R}$ with exactly two equivalence classes.  Each equivalence class should be dense in $\mathbb{R}$. There should be some kind of symmetry between the two equivalence classes. (This is intentionally imprecise.)   If we divide $\mathbb{R}$ into rationals and irrationals, this lacks symmetry, since the rationals have measure zero and are countable, while the rationals have infinite measure and are uncountable.   So we could require that each equivalence class be uncountable and/or have positive (or infinite) measure. Then take one equivalence class to be the rationals, union with countably many fat cantor sets with positive measure, and the other class its complement. Then both classes are now dense, uncountable, and have infinite measure. But again, symmetry is lacking. One of these classes is "uncountable everywhere" and has "positive measure everywhere" while the other does not.   I propose some definitions: A subset $A \subset \mathbb{R}$ is uncountable everywhere if for any open interval $(a,b)$, the intersection $A \cap (a,b)$ is uncountable. A subset $A \subset \mathbb{R}$ has positive measure everywhere if for every open interval $(a,b)$, the intersection $A \cap (a,b)$ has positive measure. We can see immediately that having positive measure everywhere implies being uncountable everywhere, since a positive measure set must be uncountable.   Finally, my questions:    Is there a partition of $\mathbb{R}$ into two sets that are both uncountable everywhere?  Is there a partition of $\mathbb{R}$ into two sets that both have positive measure everywhere?  Is there a partition of $\mathbb{R}$ into two sets that split each interval $(a,b)$ into two parts of equal measure?   Potential generalizations for extra credit: What about partitions of $\mathbb{R}$ with $n$ equivalence classes, where $n \in \mathbb{N}$, or even with countably many, or even uncountably many equivalence classes?  Note: If you've seen these definitions of uncountable everywhere or positive measure everywhere somewhere under a different name, please let me know. I've never found anything where other people were thinking about these notions.  EDIT: By "measure," I mean Lebesgue outer measure, so we don't have to worry about anything being measurable. i know that|$S_n$|=$n!$ and |$A_n$|= $\frac{n!}{2}$ but i'm not sure where to go from there. Thanks. Prove that If $m'$ is a common multiple of $s$ and $t$, then $m | m'$. Here $m$ is the LCM of $s$ and $t$.    Although the statement is intuitively clear to me I don't know how to prove. $f'(x) = [f(x)]^{2}$. Prove $f(x) = 0 $ Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f(0) = 0$ and $f'(x) = [f(x)]^{2}$, $\forall x \in \mathbb{R}$. Show that $f(x) = 0$,  $\forall x \in \mathbb{R} $.  I first (unsuccessfully) tried using the Mean Value Theorem, but this is in the Integrals chapter so the solution probably involves them. Can't really see where integrals come in here though.  What I've got so far:   (i) Since $f$ is differentiable, thus it is continuous and, hence,  integrable. Therefore $f^2$ is also integrable and as $f'=f^2$, $f'$ is too.   (ii) $f' \geq 0$, $\forall x \in \mathbb{R}$ USB Charge and data separate cables Is it possible to split micro USB to USB cable so two cables (data + and data -) go to the PC with USB and two other power cables go to power brick? Is there an idiom for someone who gets what he deserved? Like someone receiving punishment for his evil deeds or someone getting awarded for his good deeds? How do I set focus follows mouse? Is there a way to set up "focus follows mouse" behavior in Unity? Elements order in a "for (… in …)" loop Does the "for…in" loop in Javascript loop through the hashtables/elements in the order they are declared? Is there a browser which doesn't do it in order? The object I wish to use will be declared once and will never be modified.  Suppose I have:  var myObject = { A: "Hello", B: "World" };   And I further use them in:  for (var item in myObject) alert(item + " : " + myObject[item]);   Can I expect 'A : "Hello"' to always come before 'B : "World"' in most decent browsers? Checking from shell script if a directory contains files From a shell script, how do I check if a directory contains files?  Something similar to this  if [ -e /some/dir/* ]; then echo "huzzah"; fi;   but which works if the directory contains one or several files (the above one only works with exactly 0 or 1 files). I was told that the following does not work:     Employees having children will be granted a day off.   However, this sounds better:     Having children, those employees will be granted a day off.   What is wrong with the first clause? What is the inverse function of $e^x +x$? Prove the equality: $\det\left[\begin{smallmatrix} -2a &a+b &a+c \\ b+a&-2b &b+c \\ c+a&c+b &-2c \end{smallmatrix}\right] = 4(a+b)(b+c)(c+a)$ i have to prove:   $$\begin{vmatrix} -2a &amp;a+b  &amp;a+c \\   b+a&amp;-2b  &amp;b+c \\   c+a&amp;c+b  &amp;-2c  \end{vmatrix} = 4(a+b)(b+c)(c+a)$$   I have tried many calculations between the rows and columns of the determinant to get to the answer i want to solve the exercise however none of them gave me something right. Can someone help? Which kind of pointer do I use when? I'm trying to draw a rounded rectangle with round "holes" using  but I need the rectangle borders to be thick and "holes" borders to be thin. Is there a better way to do so than:  \begin{tikzpicture}[x=1mm, y=1mm]     % Background         \fill[fill=black!10] (1.5, 1.5) rectangle (15, -15);      % Fill     \fill[fill=black!2, even odd rule, rounded corners=0.5mm]          (0, 0) rectangle (10, -10)          (2, -2) circle[radius=1.25mm]         (2, -5) circle[radius=1.25mm];     % Rectangle border     \draw[thick, rounded corners=0.5mm] (0, 0) rectangle (10, -10);     % Holes borders     \draw[thin] (2, -2) circle (1.25mm);     \draw[thin] (2, -5) circle (1.25mm); \end{tikzpicture}   The result image should look like this: Is Java "pass-by-reference" or "pass-by-value"? Spatial query to return multiple values I have a woodland scheme advice layer that has a polygon for each area of advice (brown layer in the picture). I also have a polygon layer of UK local authorities (red boundary layer in the picture). I would like to perform a spatial query which returns the local authority name for each woodland advice scheme. This would work fine if each woodland advice scheme was found in just one local authority boundary. However many are found across a boundary of two different authorities. For instance, one scheme sits across the boundaries of the authorities of Fife and Angus. The scheme is the yellow selected polygon in the picture and the red line is the authority boundary. How do I get the spatial join to return a value of Fife; Angus (or something similar to identify two values) instead of just Fife or Angus. How could a gap between array modifier created objects be closed? If I have a Java source file (*.java) or a class file (*.class), how can I convert it to a .exe file?  I also need an installer for my program. How does the coin limit work?