How to find $\sum_{n=0}^{\infty}\frac{1}{2^{2^n}}$? How can we handle sums of the type:$$\sum_{n=0}^{\infty}\frac{1}{2^{2^n}}$$ It is different from geometric series. Is there any general approach to deal such sums?
Sum of Infinite Series $1 + 1/2 + 1/4 + 1/16 + \cdots$ Everyone knows about the classic  $$ \sum_{i=1}^{\infty} \dfrac{1}{2^i} = 1 $$ However, is there any way to find $$ \sum_{i=0}^{\infty} \dfrac{1}{2^{2^i}} = \dfrac12 + \dfrac14 + \dfrac{1}{16} + \dfrac{1}{256} + \cdots $$
I've read that abelian categories can naturally be enriched in $\mathbf{Ab}.$ How does this actually work? Wikipedia defines the notion of an abelian category as follows ().     A category is abelian iff         it has a zero object,   it has all binary products and binary coproducts, and   it has all kernels and cokernels.   all monomorphisms and epimorphisms are normal.      It later explains that an abelian category can naturally be enriched in $\mathbf{Ab},$ as a result of the first three axioms above.  How does this actually work? What happens if you plug more than one charger in the new MacBook Pro (2016)? Apple's  decision to convert all the ports on their 2016 MacBook Pro lineup to USB-C ports contributes to blur the difference between ports that supply power (such as a power adapters) and ports that can power external devices (such as USB keys, mice or other gadgets).    This made me think about the following scenario: you sit at a table where your laptop is plugged in on the right, and don't notice that it is plugged in, so you decide to charge it with a second nearby charger that is laying around (i.e. belonging to someone who lives with you and who recently sat at the same spot).  What would happen then?   Would the computer charge twice as fast, would one of the chargers be dismissed by the system (which one?), or would it generate a surcharge that would destroy the computer?  I couldn't find anything about this online, but since I am always trying to force a USB in the MagSafe charger of my MacBook Air, I think that I could easily end up making this kind of mistake.  Note: My question asks what happens when two chargers are inserted into a MacBook Pro, but I am also wondering what would happend if 3, or 4 chargers were inserted. Although this scenario is unlikely to happen by mistake, it would constitute an obvious vulnerability to an exposed device, i.e. in a public space. Normal subgroups of $S_4$ Can anyone tell me how to find all normal subgroups of the symmetric group $S_4$?  In particular are $H=\{e,(1 2)(3 4)\}$ and $K=\{e,(1 2)(3 4), (1 3)(2 4),(1 4)(2 3)\}$ normal subgroups? Can't get my model out of wireframe mode What are some examples of awkward sounding but grammatically correct sentences? What are some examples of awkward sounding but grammatically correct sentences? True displacement in EEVEE? I have a Blender 2.8 scene with animated true displacement. It is created using material nodes and works fine in Cycles. However, EEVEE doesn't displace my meshes truly, only normals are changed. Is it possible and how to solve the problem now or in the future? How to install the latest versions of NodeJS and NPM? Finding my house in minecraft I wandered too far to go look for sheep and now I'm on a completely different island.  How do I find my way back to my house? How do I lerp between values that loop (such as hue or rotation)? I'm trying to make the joint rotate smoothly around the center of the canvas, toward the angle of mouse pointer. What I have works, but I want it to animate the shortest distance possible to get to the mouse angle. The problem occurs when the value loops around at the horizontal line (3.14 and -3.14). Mouseover that area to see how the direction switches and it takes the long way back around.  Relevant Code   // ease the current angle to the target angle joint.angle += ( joint.targetAngle - joint.angle ) * 0.1;  // get angle from joint to mouse var dx = e.clientX - joint.x,     dy = e.clientY - joint.y;   joint.targetAngle = Math.atan2( dy, dx );   How can I make it rotate the shortest distance, even "across the gap"? Prove that the sequence with $T(0)=1$ and $T(n) = 1 + \sum_{j=0}^{n-1}T(j)$ is given by $T(n)=2^n$ $T(0)=1 \\ T(n) = 1 + \sum_{j=0}^{n-1}T(j) \\ $  Show that $T(n) = 2^n$.  I know how to prove this by induction, but I would like to know how to show this using first principles.  Edit: The way I want to solve this problem is manipulate $T(n)$ in such a way that it ends up as $2^n$. Tensor product of a module with an ideal is isomorphic to their standard product Is there a way to log all Microsoft SQL Server queries executed on a certain server, with their respective durations?  I'd like to be able to analyze all queries will a tool like what  does for PostgreSQL. While we're at it: do you know of any such tools? It seems that through various related questions here, there is consensus that the "95%" part of what we call a "95% confidence interval" refers to the fact that if we were to exactly replicate our sampling and CI-computation procedures  many times, 95% of thusly computed CIs would contain the population mean. It also seems to be the consensus that this definition does not permit one to conclude from a single 95%CI that there is a 95% chance that the mean falls somewhere within the CI. However, I don't understand how the former doesn't imply the latter insofar as, having imagined many CIs 95% of which contain the population mean, shouldn't our uncertainty (with regards to whether our actually-computed CI contains the population mean or not) force us to use the base-rate of the imagined cases (95%) as our estimate of the probability that our actual case contains the CI?   I've seen posts argue along the lines of "the actually-computed CI either contains the population mean or it doesn't, so its probability is either 1 or 0", but this seems to imply a strange definition of probability that is dependent on unknown states (i.e. a friend flips fair coin, hides the result, and I am disallowed from saying there is a 50% chance that it's heads).  Surely I'm wrong, but I don't see where my logic has gone awry... How can I make focus follow the mouse cursor? 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"? The differences between TeX engines I accidentally issued some commands I thought were safe in the terminal, and I ended up with no System Settings and lots of other missing functionality, but the OS still boots normally. I've decided I want to install Lubuntu instead of fixing Ubuntu for a fresh start and since my laptop is running on a 1GHz dual core AMD CPU.  I created a bootable USB as I usually do and placed Lubuntu 17.10 amd64 using Rufus 2.14.1086. I have GRUB that was installed with Ubuntu upon first install I still have Win8.1 which I still want to keep.  PS: I also tried creating a startup disk from Ubuntu  The question is: How can I boot from USB with GRUB2 so I can overwrite Ubuntu? If Chrome is already open, it opens new sites (e.g. from Windows shortcuts, application shortcuts, or from the Run dialog) in new tabs of the existing instance.   I'd prefer to open new sites in a separate window (a new instance of Chrome).   I don't see that in the options - is it possible? How do I backup and restore SMS Messages? How do I backup and restore SMS Messages? Prove that if the sum of each row of $A$ equals $s$, then $s$ is an eigenvalue of $A$. Let $A$ be an $n \times n$ matrix.  $i)$Prove that if the sum of each row of $A$ equals $s$, then $s$ is an eigenvalue of $A$.  $ii)$Prove that if the sum of each column of $A$ equals $s$, then $s$ is an eigenvalue of $A$.  I think that being an eigenvalue of $A$ implies that $sv=Av$ for some vector $v$. Furthermore, I know that $[a_i] = s$ if we let $a_i$ denote the i-th row of $A$. However, I do not seem to be able to find a link between these two facts. Could anyone please help me out?