Let $(x_n)$ be a real-valued sequence with partial sums $s_n = x_1 + x_2 + ... +x_n$. We define $\sigma_n = \frac{1}{n}(s_1 + s_2 + ... + s_n)$. Now, supposing $\Sigma\,x_n$ is convergent, I need to show that the sequence $(\sigma_n)$ is convergent and that $lim\,\sigma_n = \Sigma\,x_n$.  My intuition is to simply say that since $\Sigma\,x_n$ converges, then $lim\,sup\,s_n = lim\,s_n = L$, some real number, and then try to find a way to use a convergence test.   But I don't have any information about the $x$-values, so I can't say they're non-negative, which means I can't use comparison, and the ratio and root test both come out inconclusive since $lim\,\frac{\sigma_{n+1}}{\sigma_n}=1$.  Should I be finding a way to use the Cauchy criterion here? Or am I missing a different technique altogether?   EDIT: This isn't actually a duplicate of the , because there is only one series of partial sums in the arithmetic mean. This problem takes the sum of each of those sums from 1 to n and presents a new problem with that.  EDIT 2: Upon further review, although not identical, the problem turns out to essentially be a re-skin of the arithmetic mean. See answer below.
Let $(a_n)_{n \in \mathbb{N}}$ be a convergent sequence with limit $a \in \mathbb{R}$. Show that the arithmetic mean given by: $$s_n:= \frac{1}{n}\sum_{i=1}^n a_i \tag{A.M.} $$ also converges to $a$.  I have read:  but unfortunately the answers there don't help me much because I don't understand their substitutions and most of it all, why their substitutions seem to work.   What I know from the problem is that since $a_n$ is convergent and $\epsilon &gt;0$ is given, I can say that:   $$\exists N_1 \in \mathbb{N}, \forall n \geq N_1: |a_n-a|&lt;\epsilon_1 $$ I also know that since $a_n$ is convergent, it is bound, so $(a_n) &lt; M, \ \forall n \in \mathbb{N}$   I need to show that: $$\exists N_2 \in \mathbb{N}, \forall n \geq N_2: |s_n-a|&lt; \epsilon_2$$ I started as follows: $$ \left|s_n-a \right| = \left|\frac{1}{n}\sum_{i=1}^na_i -a\right|= \left|\frac{1}{n}\sum_{i=1}^m(a_i-a)+\frac{1}{n}\sum_{i=m+1}^n(a_i-a) \right| \\ \leq \frac{1}{n} \sum_{i=1}^m|(a_i-a)|+\frac{1}{n}\sum_{i=m+1}^n|a_i-a|$$ I believe to understand that the left sum after the $\leq$ is finite, bound and doesn't depend on the values that $n$ takes on. However, I don't understand where all the substitutions come from and make this proof so seemingly easy to complete.   Is there a general idea I can follow to complete such proofs? Because I know that the last step is to show that the given sum is smaller than $\epsilon$. I also know that I should bring the condition $a_n &lt; M$ into place somewhere, but I don't know where.  If I choose an $n \geq N: |a_n-a|&lt; \epsilon'$ what does that tell me about $|a_i-a|$? I know that they are seemingly the same just with a different index.
Basically I want  \mint{python}|'banana'|   to produce straight single quotes, rather than the curly ones it normally produces (in pdflatex). How was Thor able to dent Captain America's Shield? I was reading about Adamantium on the      Proto-Adamantium   This is the original Adamantium created by Dr. MacLain. Proto-Adamantium is the official name for the metal that makes up Captain America's Shield. It has never been recreated and is the absolute most invulnerable item in all of existence.    This lead me to read about      The true shield was once slightly dented by an enraged Odinforce empowered Thor, using Mjolnir, but he later reconciled and pounded out the small dent.    Was the dent caused by Odinforce or Thor's Hammer? Is Mjolnir or Captain America's Shield "the absolute most invulnerable item in all of existence"? Activity has leaked window com.android.internal.policy.impl.PhoneWindow$DecorView@44f72ff0 that was originally added here Let's say we have $x^2=25$ So we have two real roots ie $+5$ and $-5$.  But if we were to differentiate on both sides with respect to $x$ we'll have the equation $2x=0$ which gives us the only root as $x=0$.  So does differentiating on both sides of an equation alter it? If it does, then how do we conveniently do it in Integration by substitutions? If not then what exactly is going on here ? I'm looking at buying a second-hand Mac, probably one that does not include an OS installation disk.  How can I scan it to make sure that it has not had security threats installed on it?  Is there a CD or DVD image I can boot it up with to perform this scan offline?  Edit: prospective machine would be a 2011- or 2012-vintage MacBook Pro.  Also note that I don't have reliable access to high-speed internet, so solutions that involve downloading 5GB over the Internet are not ideal.  For others, though,  may be helpful. ffmpeg working from command line in Terminal but not in an Automator shell script! Finding all solutions for a system of equations with constraint on the sum of absolute values I am looking for a fast algorithm to find all integer solutions for the following system of linear equations ($c_1,\ldots,c_n \in \mathbb{Z}$ and $r\in \mathbb{N}$ ):  $x_1+x_2=c_1$  $x_3+x_4=c_2$  ...  $x_{2n-1}+x_{2n}=c_n$  $|x_1|+|x_2|+\ldots+|x_{2n}|=r$  $x_1,\ldots,x_{2n} \in \mathbb{Z}$   Also can we find the number of solutions before solving the system? How to convert SSH public key from PEM to DER format? I'd like to convert an existing public key (generated using ssh-keygen) from PEM-format to DER-format  I tried:  openssl rsa -in ~/.ssh/id_rsa.pub -inform PEM -pubin -outform DER -out ~/.ssh/id_rsa.pub.der   Unfortunately this gives the following error:  unable to load Public Key 139946985690784:error:0906D06C:PEM routines:PEM_read_bio:no start line:pem_lib.c:703:Expecting: PUBLIC KEY   The file ~/.ssh/id_rsa.pub contains a public key; cat ~/.ssh/id_rsa.pub gives:  ssh-rsa ...... user@machine   What am I doing wrong? I am wondering if my proof is correct? Thank you for whoever willing to take a look at it for me.     Proof $\liminf E_k \subset \limsup E_k $   If $\{E_k\}_{k=1}^\infty$ is a sequence of sets, we define \begin{align*} \limsup E_k &amp; = \bigcap_{j=1}^\infty\left(\bigcup_{k=j}^\infty E_k\right)\\ &amp;= \bigcap_{j=1}^\infty(E_j \cup E_{j+1} \cup \cdots)\\ &amp;= (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots. \end{align*}  Therefore, $\limsup E_k$ consists of those points in $\mathbb{R}^n$ which belong to infinitely many $E_k$. \begin{align*} \liminf E_k &amp; = \bigcup_{j=1}^\infty\left(\bigcap_{k=j}^\infty E_k\right)\\ &amp;= \bigcup_{j=1}^\infty(E_j \cap E_{j+1} \cap \cdots)\\ &amp;= (E_1 \cap E_2 \cap \cdots) \cup (E_2 \cap E_3 \cap \cdots) \cup \cdots. \end{align*} Therefore, $\liminf E_k$ consists of those points in $\mathbb{R}^n$ which belong to all $E_k$ for $k \geq k_0$.  Now consider $x \in \liminf E_k$, then $x \in E_1 \cap E_2 \cap \cdots$, or $x \in E_2 \cap E_3 \cap \cdots$ and so on, that is to say, $x \in E_{k_0} \cup E_{k_0+1} \cup \dots$ for some $k_0$. More specifically, $x \in E_k \forall k \geq k_0$. Therefore, $x \in (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots$, so we showed $x \in \limsup E_k$. Let A and B be sets. Show that A is a subset of B if and only if for any set C, one has A union C is a subset of B union C. Can you verify my proof if it is right?  Let $A$ and $B$ be sets.  (a) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ union $C$ is a subset of $B$ union $C$. (b) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ intersect $C$ is a subset of $B$ intersect $C$.  Edited Solution :  (a)  (=>) Given A is a subset of B, this implies A U B = B. Given a set C, A U B U C = B U C. Since A is a subset of B, this implies A U C is a subset of B U C.      Solution 1 :(&lt;=) Given A U C is a subset of B U C , For A U C , x is a member of A U C . For B U C, x is a member of B U C. Suppose x is a member of A but not a member of B, meaning A is not a subset of B. Therefore , A U C is not a subset of B U C. Hence, A must be a subset of B.      Solution 2 :(&lt;=) Given A U C is a subset of B U C, For A U C, x is a member of A U C and for B U C, x is a member of B U C. Suppose C is a disjoint set. Since A U C is a subset of B U C, it follows that A is a subset of B.  (b) (=>) Since A is a subset of B , it follows that A U B = B. [A intersect C] is a subset of A and  [B intersect C ] is a subset of B by definition of intersection. Let x be member of A and member of C. Since [ A intersect C ] is a subset of A, x must be in A. Let x be a member of B and member of C. Since [ B intersect C ] is a subset of B, x must be in B. Therefore, A intersect C is a subset of B intersect C.  (&lt;=) Since A intersect C is a subset of B intersect C, suppose we let Set C be a subset of B and a subset of A as well. C is a subset of A implies A intersect C is equals to C. C is a subset of B implies B intersect C is equals to C as well. Since A intersect C is equals to B intersect C, A is a subset of B. Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice? Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice?  What would it imply if there was such an example?  EDIT: For instance, maybe this will help understand the kind of example I had in mind, it is known that an important feature to determine that the AC is needed in the Banach-Tarski case is the non-transitivity of rotations on Euclidean space, one might find an example of a transitive group for some given space keeping the rest equal and make the AC unnecessary?  I guess then it might be said that this transitivity will be in this particular case equivalent to the AC or some amount of it, but I guess that it would be important to show this if it hadn't been realized before. Show that $\dim H^0(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(m)) = {n + m \choose n}$ if $m \geq 0$, and $0$ otherwise. Close and flag links do nothing on mobile site This morning (it's about six AM here) I find that I can no longer submit close votes; the link just jumps to the beginning of the page.  I'm on iOS, using a third-party browser called Grazing, but a brief check with Mobile Safari shows the same problem there.   Here on Meta, where my score is much lower, I only get a "delete" link on my own question, and that works fine. (Oops :-))   Update: As per comments, this affects the "flag" link, too, which doesn't even scroll to the top of the page. Update 2: Latest comment suggests this affects Android, too. Update 3: This also affects desktop computers when viewing mobile (when you click "mobile" in the footer) Update 4:  mentions Windows 8, too; simply deleting platform-specific tags Most of you if not everybody will agree that the stronger the gravitational pull, the faster an object will fall. For example, on a planet with 50 times the gravity of Earth, any object will hit the ground on that planet much quicker than it would on Earth. So taking all of these into the equation, does it mean that at a blackhole, an object will fall at an infinite speed because of the infinitely strong gravitational pull of the blackhole? For which real values of t does the following system of linear equations:  $$ \left\{  \begin{array}{c} tx_1 + x_2 + x_3 = 1 \\  x_1 + tx_2 + x_3 = 1 \\  x_1 + x_2 + tx_3 = 1 \end{array} \right.  $$  Have:  a) a unique solution?  b) infinitely many solutions?  c) no solutions?  I haven't done linear algebra in almost a year, so I'm really rusty and could use some pushes in the right direction. A set of elements in a reduced unity ring Let $(A,+,\cdot)$ be a unity ring with the property that if $x \in A$ and $x^2=0$ then $x=0$. Consider the set $M=\{a\in A | a^3=a\}$. Prove that:      a) $2a\in Z(A)$, $\forall a\in M$, where $Z(A)$ denotes the centre of the ring $A$;   b) $ab=ba$, $\forall a,b\in M$.   My attempts revolved around the fact that an idempotent element in a reduced ring is central. So, since for $a\in M$ we have that $(a^2)^2=a^2$, it follows that $a^2\in Z(A)$, $\forall a\in M$. The next thing I wanted to use in order to solve a) was that $Z(A)$ is a subring of $A$, so if I had proved that $(a+1)^2 \in Z(A)$, $\forall a\in M$, then we would have reached the desired conclusion. However, I couldn't prove this and I honestly doubt that it is true. Another idea that I had was to prove that $M$ is a subring of $A$. Of course, this didn't work out because I cannot even prove that $M$ is closed under addition. Again, I don't know if this is true and it most likely isn't. As for b), I think that a) should be of use, but I don't know how. It is a well-known problem that a ring with $x^3=x$ for any $x$ in that ring is commutative, but since $(M,+,\cdot)$ is almost definitely not a ring, this doesn't help. EDIT: Is there any chance that this question is simply wrong? I tended to believe this before asking it here too, but since nobody has made any progress on it until now I am even more inclined to think so. How to change inside background color of UISearchBar component on iOS I'm trying to compute the shadow of 3D objects on the ground. To do this I'm assuming parallel rays from Sun origin.   Let's assume the Sun direction is given by azimuth $\varphi$ and elevation $\theta$ angles. So the direction unit vector is retrieved as:  $$ \hat{\mathbf{s}} = \mathbf{R}_y\left(\theta\right)\mathbf{R}_z\left(\varphi\right)\hat{\boldsymbol{\imath}}$$  where $\mathbf{R}_{x/y/z}$ are the rotation matrices along the $x, y, z$ axes given by, for example, Euler-Rodrigues equation, and $\hat{\boldsymbol{\imath}}$ is the unit vector along $x$-axis.  A point on a ray-line has the equation:  \begin{equation} \mathbf{r} = \mathbf{r}_p + \lambda\hat{\mathbf{s}}  \tag{1} \end{equation}  being $\mathbf{r}_p$ a point on that line.  A point on a general plane is given by the equation:  \begin{equation} \left(\mathbf{r} - \mathbf{r_0}\right)\cdot \hat{\mathbf{n}} = 0 \tag{2} \end{equation}  being $\mathbf{r}_0$ a point on the plane and $\hat{\mathbf{n}}$ the unit vector normal to that plane.   Ensuring the validity of both equations (1) and (2) allows to compute the $\lambda$ multiplier and retrieve the position vector of the projected point (writing $\tilde{\mathbf{r}} = \mathbf{r_p} - \mathbf{r_o}$)  \begin{gather} \lambda = - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}} {\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right) \\ \mathbf{r}_\text{proj} = \mathbf{r}_p - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}} {\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right)\hat{\mathbf{s}} \tag{3} \end{gather}  Applying eq. (3) to all the vertices of a mesh $\left\{\mathbf{r}_{p_i}\right\}$ .   A problem can be noted: $\hat{\mathbf{s}}\cdot \hat{\mathbf{n}}$ can be zero (e.g. Sun is horizontal, i.e. the ray is ortogonal to the normal of plane).   Are there some conceptual mistakes in my reasoning? Is there a way to rewrite the operation as a linear operator applied to the set of vertices position vectors?   $$ \left[\mathbf{A}\right]\left\{\mathbf{r}_{p_i}\right\} = \mathbf{r}_\text{proj}$$ How to enable virtualization without using bios? I am using android studio.I want to run my app on simulator but it is asking me to install intel haxm,.When i install intel haxm it says that vitualization is not enabled/.To enable virtualization i need to goto bios but issue my bios is password protected &amp; i don't know password.So please tell me is there any other way by which i can enable the virtualization without going into bios? How can we make First letter Capital while using trigonometric functions. eg: \cos (60^\circ) gives simple cos (all in small). What is needed is 'Cos' (with initial capital). For italics in mactex i used \it but it didn't italicized 'cos' only the number '60'. So how to do that?