Use Mathematical induction to prove that for all integers, $n$ is greater than or equal to $1$. I am confused on what to do after I do the the basis step that is using $n$ as $1$.   $$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} +...+\frac{1}{n(n+1)}=1-\frac{1}{n+1}$$
I am trying to prove the summation formula using induction:  $$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$$  So far I have...  Base case:   Let n=1 and test   $\frac{1}{k(k+1)} = 1-\frac{1}{n+1}$  $\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$    $\frac{1}{2} = \frac{1}{2}$   True for n=1   Induction Hypothesis:   Assume the statement is true for the n-th case   $\sum_{k=1}^n  \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$  Inductive Step:   Prove, using the Inductive Hypothesis as a premise, that   $$\sum_{k=1}^{n+1}  \frac{1}{k(k+1)} = \sum_{k=1}^{n}  \frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)} = 1-\frac1{n+1} + \frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)}{(n+1)(n+2)}+\frac{-2-n}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)-2-n+1)}{(n+1)(n+2)} = \frac{(n+1)(n+2)-n-1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{(n+1)(n+1)}{(n+1)(n+2)} = \frac{n+1}{n+2}$$  To prove $$ 1-\frac{1}{n+2} = \frac{n+1}{n+2} $$ Multiply both sides by $n+2$ to get an equivalent expression. $$ (1-\frac{1}{n+2}) * (n+2) = (\frac{n+1}{n+2}) * (n+2)  $$ $$ n+1=n+2−1 $$  Does this all make sense? How can this be improved upon?
I am trying to customize the legend for this graph. First of all, I would like the legend width to be equal to the x-axis width. At the moment, I am setting the width of the legend via      /tikz/every even column/.append style={column sep=1.0cm}   and I could manually adjust this value so that length equals the x-axis width, but I guess that there is a more elegant way to set the width of the legend.  Second, I don't really like the little color rectangles in the legend. I only want one rectangle in front of every entry and not two. This rectangle does not need to have a border and should be vertically centred with the text. The alignment seems to be okay for the rectangle closer to the text, but not for one the far left.   Update: Thanks to Christian Feuersänger's comment, there is possibility to change the rectangles via   legend image code/.code={%     \draw[#1, draw=none] (0cm,-0.1cm) rectangle (0.6cm,0.1cm);   or what Gonzalo Medina proposed  legend image code/.code={%     \draw[#1, draw=none, /tikz/.cd, bar width=3pt, yshift=-0.2em, bar shift=0pt] %     plot coordinates {(2*\pgfplotbarwidth,0.6em)};   Third, I would like to the legend entries to start next to the text "Legend:".  I don't really know if with these requirements it still makes sense to use legend or if it would be better to add everything manually via after end axis/.code and \ref{label}.  Graph    MWE  \documentclass{article} \usepackage{pgfplots} \pgfplotsset{compat=newest}  \begin{document} \begin{tikzpicture}     \begin{axis}[                 ybar,                 height=6cm,                 title={Some plot},                 width=12cm,                 axis on top,                 y axis line style={opacity=0},                 axis x line*=bottom,                 tick align=inside,                 tick style={draw=none},                 xtick=data,                 ymajorgrids,                 major grid style={draw=white},                 xtick={1,2,3,4},                 legend style={                     at={(0.0,-0.15)},                     anchor=north west,                     legend columns=-1,                     /tikz/every even column/.append style={column sep=1.0cm}                         },             ]                    \addlegendimage{empty legend}             \addlegendentry{\textbf{Legend:}}                                \addplot coordinates {                 (1, 0.2064)                 (2, 0.3961)                 (3, 0.5961)                 (4, 0.7961)}; \addlegendentry[text width=45pt, text depth=]{legend a}             \addplot coordinates {                 (1, 0.3064)                 (2, 0.4961)                 (3, 0.6961)                 (4, 0.8961)}; \addlegendentry[text width=65pt, text depth=]{lengend b}     \end{axis}  \end{tikzpicture} \end{document} Most common C# bitwise operations on enums For the life of me, I can't remember how to set, delete, toggle or test a bit in a bitfield. Either I'm unsure or I mix them up because I rarely need these. So a "bit-cheat-sheet" would be nice to have.  For example:   flags = flags | FlagsEnum.Bit4;  // Set bit 4.   or  if ((flags &amp; FlagsEnum.Bit4)) == FlagsEnum.Bit4) // Is there a less verbose way?   Can you give examples of all the other common operations, preferably in C# syntax using a [Flags] enum? I have been given the task to rewrite a function as a sum of an even and an odd function. But when I went to analyze the parts of the original function I noticed that one part was neither even nor odd.   The original function was: $g(x) = (x + 1) / (x^2 - 3x + 4$)  After analyzing I realized that when using f(-x) to determine symmetry it produced a 'neither' result. How would I go about writing a function that is neither even nor odd as an even function. (It would have to be an even function as the (x+1) part of the original function is odd and since I need to write it as a sum of even and odd, the denominator would need to be even. Pass commands as input to another command (su, ssh, sh, etc) Why doesn't grep using pipe work here? I have the following command:  find / -name libGL.so.1   Which returns lots of lines with "Permission denied". I want to exclude such lines, so I added the following:  find / -name libGL.so.1 | grep -v 'denied'   But the output is the same - my grep -v 'denied' is not filtering out the lines with Permission denied. I've tried many variations, looked over grep tutorials, but I cannot figure out the problem. Any suggestions? Usage of “when” or “while” in this context He left the house _ I was still sleeping   Would this sentence be better completed with when or with while? Just don't show UI for actions that can't be actioned Recently I just deleted one of my answer in stackoverflow where there was one comment. After deleting the answer it also show for sometime that we all know but in the comment it still allow us to up to make it useful but as I clicked it it just showed me the message:  If we are not allowed to do then what was the meaning? Why don't it just be removed or make it disable if the answer is deleted. Just don't show UI for actions that can't be actioned Selection by attribute string with wildcard as part of name in ArcGIS Desktop? I have an attribute field in my file geodatabase (GDB) where I'm trying to select all records that contain an underbar in the string field.  The issue is that the underbar is considered a SQL wildcard on its own.  To illustrate: I need all data similar to "ID" = '42W_42_42E' or "ID" = '31A_31'  Using "ID" LIKE '%_%' returns all records, which isn't helpful.  The perfect solution would simply return a count on the number of records with the underbar in their ID attribute. Currently, when flagging an off-topic question, the following choices are offered.   Questions about general computing hardware and software are off-topic for Stack Overflow. Questions concerning problems with code you've written must describe the specific problem and include valid code to reproduce it. Questions must demonstrate a minimal understanding of the problem being solved. Questions on professional server, networking, or related infrastructure administration are off-topic for Stack Overflow unless they directly involve programming or programming tools. Questions asking us to recommend a tool, library or favorite off-site resource are off-topic for Stack Overflow as they tend to attract opinionated answers and spam. This question belongs on another site in the Stack Exchange network   However, there are some questions that fit in none of these categories.  For example, career advice questions are off-topic here (and on programmers.SE), questions like  are off-topic on pretty much any SE site.    Some questions might also be relevant on another SE site not listed when the "belongs on another site" option is selected.  How should those questions be flagged?  Why isn't there a "it's off-topic because it is" option?  EDIT: just to be clear, I don't have the "Other" option under Off-Topic. Insect identification - Is this a bedbug? This bug was found on a couch in . The couch, and the two beds in the room show no signs of other bugs. Is this a bed bug that I should worry about or something harmless? It was slightly larger than a common tick. Understanding the proof for showing that there are infinitely many $n\in\mathbb{Z}^+$ such that $n!+1$ is divisible by at least two distinct primes There is another post pertaining to this question; however, the other post does not address some specifics that concern me in the proof. Proof: Let $n=p-1, p \geq 7$. Then $p|(p-1)!+1$. We will show that $(p-1)!+1=p^n$ has no solutions. Suppose not. Suppose $(p-1)! + 1 = p^n$ has a solution for $p\geq 7$. Then $(p-1)! =p^n-1  = (p-1)(p^{n-1}+\cdots+p+1)$. So $(p-2)!\equiv n (\mod p-1)$. Now suppose that $(p-1)|n$. Then $p-1\leq n$. So $(p-1)!+1&lt;p^{p-1}\leq p^n$. Hence, there are no solutions. It remains to show that $(p-1)|(p-2)!$. But $p-1=2m$ means $2m|(p-2)!$. QED  Questions:  My understanding is that at the start of the proof, Wilson's theorem is used to show that at least one prime divides $n!+1$.  Why do we want to show that $(p-1)!+1=p^n$ has no solutions? The way I see it is that $p^n$ is composite, and we are basically saying that $p^n\nmid (p-1)!+1$, so no composite number divides $n! +1$. But how and where in the proof do we actually show that another prime $q \neq p$ divides $n!+1$?    Thanks in advance. Show that $13$ divides $2^{70} + 3^{70}$.   My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat's little theorem. So how can I separate the two?  Thanks! Removing the password from a PDF file I have a couple of ebooks as PDFs with passwords, however my ebook reader (sony prs600) doesn't seem to support PDFs with passwords.  What is the easiest of removing the password from a PDF (I know the password, which presumably helps a lot).  It's a bit annoying buying a book and then only being able to read it in front of a computer. Where's the summary of your own suggested edits? LaTeX minted Error I receive the following errors:  ! Package minted Error: You must invoke LaTeX with the -shell-escape flag. ! Package minted Error: You must have `pygmentize' installed to use this packag   in TexMaker Settings i wrote:     LaTeX : "/usr/texbin/latex" -interaction=nonstopmode %.tex -shell-escape      PdfLaTex: "/usr/texbin/pdflatex" -synctex=1 -interaction=nonstopmode   %.tex -shell-escape   But the error still doesn't vanish ... Do I need to tame my twitchy F5 finger? I just got this error message for the first time (I was getting this as the response to any request on the SE network for about the last five minutes, until now):     We're sorry...      There are an unusual number of requests coming from this IP address.   To protect our users, we can't process any more requests from this IP   address right now. We'll restore access as quickly as possible, so try   again soon. If you believe you have reached this page in error,   contact us.   As far as I know, I was not spamming SE or participating in DOS attack, and I find it unlikely that any of my office coworkers who share my IP address were doing so either.  Granted, I was pressing F5 a lot. But I've always pressed F5 a lot on SE, and have never gotten this message before. So my question is: Is it possible to get this error message simply by refreshing too much or too often? Or is that message reserved for a truly massive number of requests that are unlikely to be caused by one person refreshing a lot? (In the former case, I can't explain why I've never gotten it before; in the latter case, I can't explain why I just got it now). Solve the equation $28^x = 19^y+87^z$ in integers Bash Brace Expansion & Variables I have a find command:  find Directory/{Alpha,Bravo,Charlie} arg1 arg2   I want to replace Alpha,Bravo,Charlie with $find_dir  find Directory/{$find_dir} arg1 arg2   however the latter expands to   find Directory/{Alpha,Bravo,Charlie} arg1 arg2   rather then   find Directory/Alpha Directory/Bravo Directory/Charlie arg1 arg2   Why? It's part of a fairly complex bash script that may have more or less directories, not all which are relevant (so globbing Upload/* would not work). So if I setup three now and add another, I'll have to manually add it in later. Plus I need it run from the root of directories to keep finds outputs perspective (./Upload/Dir/file as opposed to ./Dir/file).   But using a variable would permit me to change that as needed and keep it relevant to other parts of the script. Is "Michael here" correct? Can Java's NullPointerException be changed to report on which variable was null?