Prove by induction $\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$ What would be the best way to solve this by the induction method?
My question concerns the inductive step of the proof below. I've typed out the work I've done so far. If anyone has any suggestions on how to proceed thanks in advance.  If $n\in\mathbb{N}$, then $(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^n})\ge\frac{1}{4}+\frac{1}{2^{n+1}}$.  $Proof.$ We use mathematical induction.  Base Case. For $n=1$ we see that $1-\frac{1}{2}\ge\frac{1}{4}+\frac{1}{4}$ since $\frac{1}{2}\ge\frac{1}{2}$.  Inductive Step. Let $n&gt;1$. Assume our statement is true for $n$. Now observe that  $\begin{align}(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n+1}})&amp;\ge\\ \left((1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n}})\right)(1-\frac{1}{2^{n+1}})&amp;\ge(\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$  My question is: How do we get from $\begin{align} (\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$ to $\frac{1}{4}+\frac{1}{2^{(n+1)+1}}$?  Most of what I've attempted has led to tedious calculations that do not simplify to the desired result. I've also tried making the argument that $(1-\frac{1}{2^{n+1}})$ must be $\ge$ to $\frac{1}{2}$ but then I'm stuck with $\frac{1}{8}+\frac{1}{2^{(n+1)+1}}$... Is my approach incorrect here?
Problem with literate and breaklines=true in listings package As the title says, the option breaklines=true seems to have an undesired interaction when using literate; in the following example the closing parenthesis doesn't get colorized   \documentclass{article} \usepackage{xcolor} \usepackage{listings}  \lstset{   literate=     {)}{{\textcolor{red}{)}}}{1}     {(}{{\textcolor{red}{(}}}{1},   breaklines=true, }  \begin{document}  \begin{lstlisting} () \end{lstlisting}  \end{document}     Commenting out the breaklines=true option produces the desired result. What is producing this odd behaviour and how can it be prevented? Zombies - Why do they need to feed on living animals/humans? Bootstrap Paradox in Interstellar Birthday line to get ticket in a unique setup At a movie theater, the whimsical manager announces that a free ticket will be given to the first person in line whose birthday is the same as someone in line who has already bought a ticket.  You have the option of choosing any position in the line. Assuming that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout the year (365 days year), what position in line gives you the best chance of getting free ticket? Show that the Sorgenfrey line does not have a countable basis. I am trying to understand this proof from Munkres' book which shows that the Sorgenfrey line does not have a countable basis. His proof is:  Let $\beta$ be a basis for $\mathbb{R}_l$. Choose for each $x$ an element $B_x \in \beta$ such that $x \in B_x \subset [x, x+1)$. If $y$ is not equal to $x$ then $B_y \neq B_x$ since inf($B_x$) = $x$ and $y$ = inf($B_y$). Therefore $\beta$ must be uncountable.   I am new to topology so I am probably just missing a simple point but I do not see why $\beta$ must be uncountable according to this argument. 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Entering Unicode characters in LaTeX Is there a single word for something that is simultaneously not interesting and not boring? I am looking for a word that would describe a concept or activity, such as playing a game X, that is1 neither boring nor interesting. In other words, it should mean ‘neutral’ wrt intrestingness (~ boringness) but still comparable on the scale of intrestingness to both ‘boring’ and ‘interesting’.  An example sentence would be  As far as I'm concerned, long-distance running is [insert word]; it is not boring but it's also not interesting.  Criteria2   Has to be unambiguous3, and may consist of two words at most. Has to be a single word. Should be formal in the sense that it is included in Cambridge Dictionary, or Merriam Webster, or both. Should not contain negating prefixes, such as ‘un-’.   Does such a word exist in the English language?  For readers' convenience, please state which criteria you believe your word suggestion has met. You may do so, e.g., by adding a disclaimer at the top or bottom of your answer. Simply writing (1., 2.) is sufficient.        1 Assume a ternary system of intrestingness.   2 If not all criteria can be met, you may relax  the requirements from bottom to top, in the following order: 4. -> 3.  -> 2. -> 1. Answers with the smallest number of relaxed criteria in this order will be preferred in accepting. For example, fulfilling  (1., 2., 3.) is better than achieving (1., 3., 4.). Situations where a  higher priority ordering fulfills fewer criteria than a lower priority  vector, such as (1.) vs (3.,  4.), will be assessed on a case-by-case  basis.   3 Words such as ‘unappealing’ thus even fail  criterium 1 because in a three-valued system ‘unappealing’ ~ ‘not  appealing’ ~ ‘not interesting’ can be taken to mean either ‘boring’ or  ‘neutral wrt intrestingness’. I have my mobile phone (Samsung Galaxy S2) running on ICS. Recently I activated Microsoft Exchnage ActiveSync to connect to my corporate email. It disabled all the security options like 'Pattern Unlock'. I hated it, and removed my microsoft active exchange account from the phone, but these options are still disabled. How do I get it enabled? What's the best way to develop old, exposed film? A friend of mine just found 3 rolls of exposed 135 (i.e. 35mm) film. They are at least 10 years old. He'd like to know if it is even worth taking them to the local hour photo lab but also if he should give them any special instructions. has been, have been , had been How do I get attention for one of my own questions without a good answer? I asked a question almost as soon as I got my account. I got a few decent answers, but none of them helped solve my problem. The question has long since become buried by newer questions, and there's still no accepted answer. Unfortunately, because there are quite a few answers with votes, to a casual observer it probably looks like the question was answered and I just neglected to mark an answer as accepted, and the system does not consider it "unanswered".  Or, I asked a question that was a bit more difficult to answer. After waiting for some time, I received no answers, little or no comments, and only a few views. The question does show up in the "unanswered" list, but no one has answered my question.  In either case, what's the protocol for me to try to get this question answered?  I see several options:   Post the question a second time, possibly linking to the original, to get attention for the original question. This is encouraged on sites like Reddit.  However, it's probably not appropriate here, since we're encouraged to check for duplicates and not submit a question that's already been asked by someone else, so this probably falls under that heading. Wait for some kind of wiki-like talk page to be implemented, then post a plea on the talk page for input and hope that activity on the talk page bumps up the question. Edit the question, but do nothing further on the Stack Exchange site. If I edit my post to say that I haven't accepted any of the current answers but am looking for more, then the question might get answered eventually by people browsing through old questions to find something they can answer. Post the question on other sites, but link to my original question on Stack Exchange. People will either answer here, or they'll answer on the other site and I can post the accepted answer here myself.  This might be considered spamming, so it might be better to do this without the link.      For more information, see the "" page in the . Show that the determinant of $A$ is equal to the product of its eigenvalues Show that the determinant of a matrix $A$ is equal to the product of its eigenvalues $\lambda_i$.   So I'm having a tough time figuring this one out. I know that I have to work with the characteristic polynomial of the matrix $\det(A-\lambda I)$. But, when considering an $n \times n$ matrix, I do not know how to work out the proof. Should I just use the determinant formula for any $n \times n$ matrix? I'm guessing not, because that is quite complicated. Any insights would be great.