Want $\sum_{k=1}^{n-1} g\left(\frac{k}{n}\right)$ where $n$ is odd and $g(t)=\dfrac{3^t}{3^t+3^{1/2}}$. $g(k/n) =\dfrac{3^{k/n}}{3^{k/n}+3^{1/2}}$. $\begin{array}\\ g(k/n)+g((n-k)/n) &=\dfrac{... View answer Accepted answer 37 votes So you have$n! \lt \left( \frac{n+1}{e} \right)^n \sqrt{n+1} $. I'll play and see what happens. Replacing$n$by$n-1$, we get$(n-1)! \lt \left( \frac{n}{e} \right)^{n-1} \sqrt{n} $. ... View answer Accepted answer 36 votes The binomial theorem allows you to write out the expansion of your polynomial immediately. It also allows you to answer such questions as "What is the coefficient of$x^{20}$in$(1+x)^{100}$?" Its ... View answer Accepted answer 34 votes Inverse symbolic calculator plus: https://isc.carma.newcastle.edu.au/ View answer Accepted answer 26 votes$(10a+1)^n$always ends in$1$and$(10b+6)^n$always ends in$6$so$(10a+1)^n+(10b+6)^n$always ends in$7$. View answer 26 votes Actually, you have two errors there: The minor one is that you seem to want a cube of side$2r$, since your integral goes from$-r$to$r$. The major error, as others have said, is that you are ... View answer 26 votes Here's a two-line, completely elementary proof that uses only Bernoulli's inequality: $$(1+n^{-1/2})^n \ge 1+n^{1/2} > n^{1/2}$$ so, raising to the$2/n$power,$$n^{1/n} < (1+n^{-1/2})^2 = 1 ... View answer 25 votes Map each rational$\frac{a}{b}$into the integer$2^a 3^b$. This shows that the number of rationals is at most the number of integers. If you want to handle the negative rationals, map the sign ($-1$,... View answer 24 votes This is not original with me. If we know that$e^x \ge 1+x$with equality only when$x = 0$,$e^{(x-e)/e} \ge 1 + (x-e)/e = x/e$or$e^{x/e} \ge x$or$e^{1/e} \ge x^{1/x}$with equality only if$x = ...

I'll first give a intuitive answer, then an analytic answer. Intuitively, the tangent goes in the same direction as the function, following it as closely as possible for a line. Any other line ...

The standard way (in other words, there is nothing original in what I am doing here) to get $H(x)$ is to write $H(x)G(x) = F(x)$ and get an iteration for the $c_n$. $\begin{array}\\ H(x)G(x) &=\... View answer 20 votes I usually say "sine-h", "cos-h", and "tan-h" with the "h" pronounced "aich" like the letter. Sometimes I pronounce "cosh" as a word with a long "o". I guess this qualifies as an answer, instead of ... View answer 18 votes Yes. The first difference of a polynomial of degree$d$is a polynomial of degree$d-1$. By induction, the$m$-th difference of a polynomial of degree$d$, when$m \le d$, is a polynomial of degree$...

I like the factorial base, where the integer part of a real number is written as $\sum_{i=2}^n a_i i!$ where the $a_i$ are integers such that $0 \le a_i < i$ and the fractional part is written as $\... View answer 16 votes The reason why the complex numbers are so special is that they are the end of a chain of questions of the form "How can we solve this equation?" or "What are the roots of this equation?". We start ... View answer 16 votes Vinogradov's theorem that all but a finite number of odd numbers are the sum of three primes. Not sure about the current state of Goldbach's conjecture. A theorem at about my level of math: All ... View answer 15 votes The rationals can be mapped into the lattice points$(n, m)$, which are an infinite set of isolated points in the plane. The irrationals, by any of the standard ways of mapping two reals into one (... View answer 14 votes If you have$(y_i)_{i=1}^n$, consider the mean squared difference from the$y_i$to a value$a$. This is$s(a) =\sum_{i=1}^n (y_i-a)^2 $. Manipulating this,$\begin{array}\\ s(a) &=\sum_{i=1}^...

A start, on my phone. Assume $j<k<l<m<n.$ Then j=2 or 3 because 1 makes the sum too large and 4 makes it too small. Therefore the left without 1/j is 1/2 or 2/3. You can get a tree of ...

I would argue a different way. By assumption, for all $n$, $a_n > a_{n+1}$, or $a_n \ge a_{n+1}+1$. Therefore, since $a_{n+1} \ge a_{n+2}+1$, $a_n \ge a_{n+2}+2$. Proceeding by induction, for ...

I'm going to try to make the most elementary possible proof. $\bmod 3$, $(0, \pm 1)^4 \equiv (0, 1)$ so $(0, \pm 1)^4+1 \equiv (1, 2)$ so $3 \not \mid r^4+1$. $\bmod 5$, $(0, \pm 1, \pm 2)^4 \... View answer Accepted answer 13 votes This is an example of a sieve. It is well known, and I used it over forty years in a program to (iirc) check for primality. Yes, the pattern continues. No, it cannot be used to prove the twin prime ... View answer 12 votes A Google search for "Conway and Coxeter murder" came up with the book "King of Infinite Space: Donald Coxeter, the Man Who Saved Geometry". A search via Google inside this book came up with this on ... View answer 12 votes$\sin(n) _{n=1}^{\infty}$(originally entered$\sin(1/n)$, because I was thinking of$\sin(1/x)$as$x \to 0$.) View answer 12 votes It is very well determined.$\lim_{n \to \infty} \frac{n}{n+1} = 1$(since$1-\frac{n}{n+1} = \frac{1}{n+1} \to 0$). Since$\sqrt{}$is continuous at$1$, if$f(n) \to 1$as$n \to \infty$,$\sqrt{f(...

As for the pronunciation, just remember this (not by me, but I do not know who the author is): \begin{align} &\text{A theorem both deep and profound}\\ &\text{states that "Every circle is ... View answer 11 votes Comparing naturals to integers, there is a smallest natural (0 or 1) which often makes solving problems easier. Comparing reals to complex, you can always compare reals but there is no complete ... View answer 11 votes Many questions with sum or difference of square roots can be solved with conjugating. So, ifs = \sqrt{4 + 2\sqrt{3}} - \sqrt{3}$, and$t = \sqrt{4 + 2\sqrt{3}} + \sqrt{3}$,$\begin{array}\\ st &...

In the May 2013 Fibonacci Quarterly (Vol. 51, Num. 2) pages 163-173, Michael Hirschhorn proved this result: Let $P(n) =\prod\limits_{k=0}^n \binom{n}{k}$. Then (this is going to be a pain to enter), ...
Using the suggested factorizations, and using $\begin{array}\\ k^2-k+1 &=k(k-1)+1\\ &=(k-1+1)(k-1)+1\\ &=(k-1)^2+(k-1)+1\\ \end{array}$ (this is really the key), \$\begin{array}\\ \...