What point did Southstar make that I proved? You allege that I did such a thing, so please answer that first. Otherwise you're once again just making meaningless chatter which has nothing to do with the math problems in this topic.
That has absolutely no relevance whatsoever to what both myself and SouthStar are discussing. It doesn't even make sense.
Good question. I was talking about your posts not making any sense, so why do you post things that have nothing to do with math in response to math problems?
Ok, I think I got it. I found this problem to be rather involved and tricky, took me a few different attempts at methods before I found one. Since \(\sum_{n=1}^\infty\beta_n\) converges absolutely, this means that \(\lim_{n\mapsto\infty}\beta_n=0\), so we can find a \(k\in\mathbb{N}\) such that \(\left|\beta_n\right|<\frac{1}{2}\) for \(n\geq k\). For \(n>k\), using the assumptions given in the problem, we may write: \(b_n=\frac{b_k}{\left(1+\beta_k\right)\left(1+\beta_{k+1}\right)\cdots\left(1+\beta_{n-1}\right)}\) Thus to show that the sequence \(\left{b_n\right}_{n=1}^\infty\) converges, we merely need to show that the infinite product \(\prod_{n=k}^\infty\left(1+\beta_n\right)\) converges and is strictly positive. This is equivalent to showing that the sum \(\sum_{n=k}^\infty \mbox{ln}\left(1+\beta_n\right)\) converges. For \(\left|\beta_n\right|<\frac{1}{2}\), we apply the inequality \(\left|\mbox{ln}\left(1+\beta_n\right)\right| \leq2\left|\beta_n \right|\) and, since we know that the sum \(\sum_{n=k}^\infty2\left|\beta_n\right|=2\sum_{n=k}^\infty\left|\beta_n\right|\) converges, it follows that the series of logarithms is absolutely convergent, hence convergent. Since it converges, this means that \(\prod_{n=k}^\infty\left(1+\beta_n\right)\) exists and is strictly positive. We thus conclude that \(\lim_{n\mapsto\infty}b_n\) exists.
If \(a_n\) ~ \(b_n \) as \(n \to \infty \) then \(\sum\limits_{n = 1}^\infty {a_n } \) converges absolutely iff \(\sum\limits_{n = 1}^\infty {b_n } \) converges absolutely (simple comparison test). Therefore by comments about the asymptotic behavior of ln (1+x) made earlier, \(\sum\limits_{n = 1}^\infty {\ln (1 + \beta _n )} \) converges (absolutely). But \(\sum\limits_{n = 1}^\infty {\ln (1 + \beta _n )} = \ln b_1 - {\lim }\limits_{n \to \infty } \ln b_n \)
Right... I get it now... sorry. Not the math entirely, but it seems as if you two are debating the very thing i was hinting at... Where we describe asymptotic behaviour, you cannot have any function without a reverse, so very simply, if \(\sum\limits_{n = 1}^\infty {\ln (1 + \beta _n )} = \ln b_1 - {\lim }\limits_{n \to \infty } \ln b_n \) then a single vector acting in this manner, must have a backwards directionality.... \(\sum\limits_{n = 1}^\infty {\ln (1 + \beta _n )}'' = \ln b_1 - {\lim }\limits_{n \to \infty } \ln b_n'' \) No? Because one of the first things i learned in a relativistic class, is that any singular vector has both a forward and a backward direction, a negative and a positive direction, and even in a three dimensionally-connected system, the theorem still holds true...
CptBork do you know about superior and inferior limits? There is a very interesting problem I know about which I was only able to solve recently.
Define 'backwards directionality'. 1. You've never sat a relativity class 2. Your use of the word 'singular' is wrong. 3. Relativity is irrelevent here, it's a maths problem 4. Define 'a three dimensionally-connected system' Obviously your 3 day ban did nothing to temper your lies.
If you're talking about lim sup and lim inf, I've done a few problems of that sort. Fire away Please Register or Log in to view the hidden image!
What is the smallest number of questions to be answered "yes" or "no" that one must pose in order to be sure of determining a 7-digit telephone number? If \({\{ a_n \} }\) is a positive sequence show that \( {\overline {\lim } }\limits_{n \to \infty } \left( {\frac {{1 + a_{n + 1}}}{{a_n }}} \right)^n \ge e\) and that this estimate cannot be improved upon. (The limit with the bar overhead means superior limit) I haven't definitively answered the first (I have my idea) and it took me several months (days, if you omit the 6 month break Please Register or Log in to view the hidden image!) to finally answer the second.
Point of clarification: You said 7-digit telephone number rather than 7-digit number. I assume this means the North American Plan local number, which comprises a 3 digit exchange code and a 4 digit station code. The first digit in the exchange code is 2 to 9. Can I take advantage of this in the answer? If so, 23. If not, 24.
Well, since one question that can be answered "yes" or "no" to is "Is the number larger than n?" for some given n, we can simply use a binary search of the 10^7 possibilities, giving an upper bound of \(\lceil\log_2 10^7\rceil = 24\). Of course, that's assuming that all 10^7 possibilities are represented. I'm fairly certain you can't do better with a yes/no question, as every question should partition the (remaining) numbers as evenly as possible to ensure that every sequence of answers leads to a single number as quickly as any other. Here's a cute one I heard from Peter Winckler today: Show that expanding away the parentheses of an arithmetic expression (using the four basic operations +,-,/,*) always terminates in an expression with no parentheses. This is not as trivial as it sounds. For instance, expanding \((a+b)(c-d(u+t))\) according to distribution of + gives you \(a(c-d(u+t))+b(c-d(u+t))\) which actually has more parentheses than you started out with...
Don't forget, in North America you also can't have a number starting with 911, 411, etc. Please Register or Log in to view the hidden image!
