In general your rule holds if $n$ and $m$ are coprime. Otherwise the common factor may be only one time in the prime factorisation, but is two times in the product. In your case: $2$ and $4$ are both ...

By the binomial formula we have $(1-t^2)^2=1-2 \times t^2+{(t^2)}^2=1-2t^2+t^{2 \times 2}=1-2t^2+t^4$.

It is equivalent to $\nexists n: \frac{n^2+1}{p}\in \mathbb{N}$, which is a case of the sum of squares theorem.

Its just $\frac{a}{b}-1=\frac{a}{b}-\frac{b}{b}=\frac{a-b}{b}$ as @Randall mentioned.

By the rules for logarithms we have $-\log(\frac{\sqrt{1+x^2}-1}{x})=\log(\frac{x}{\sqrt{1+x^2}-1})=\log((\frac{x}{\sqrt{1+x^2}-1}^2)^{\frac{1}{2}})=\frac{1}{2}\log(\frac{x^2}{(\sqrt{1+x^2}-1)^2})$. ...

By some easy algebraic computations, we get $x^{\log_2x}=2^{\log_2(x^{\log_2x})}=2^{\log_2x \times \log_2x}=2^{\log_2^2x}$. $\Rightarrow \log_2^2x = 4$ $\Rightarrow \log_2x = \pm 2$ $\Rightarrow ... View answer Accepted answer 2 votes As @DanielFischer pointed out, the conjecture is true:$a=b-2\Rightarrow c=(a+b)+(a*b)=(b-2+b)+((b-2)*b)=b^2-2$. Thus,$c+2=b^2$is always composite. View answer 1 votes To avoid confusion, I will call your$p$in my answer$r$. The series$\sum_{n=1}^{\infty}n^{-p}$is called$p$-series and known to be convergent iff$p>1$. Thus, your series is convergent if and ... View answer Accepted answer 1 votes Assuming that$n$is a natural number, you should divide the inequality by$n$as @RossMillikan mentioned in the comments. Then you get$\frac{a}{n-b} \leq f(a,b)$for all$n \geq 1$. Now it is easy ... View answer 1 votes The case$a_1=b_1$is trivial, so we assume$a_1>b_1$. Then by the AM-GM inequality we get$b_{n+1}<a_{n+1}$and$a_{n+2}=\frac{a_{n+1}+b_{n+1}}{2}<\frac{a_{n+1}+a_{n+1}}{2}=a_{n+1}$. View answer 1 votes$\frac{2^{-n}}{1-6^{-n}}<\frac{1}{50}$is equivalent to$2^n-3^{-n}>50$. By$3^{-n}<1$if$n>1$you get the solution. View answer 1 votes$\frac{f(12)-f(10)}{12-10}=\frac{156-110}{2}=\frac{46}{2}=23$View answer Accepted answer 1 votes The percentage difference is defined as the difference of the values, divided by the average and written in percentage. If you have two values, the formula is$\frac{|x-y|}{\frac{x+y}{2}} \times 100$. ... View answer 1 votes As @Yuval showed, the first one diverges. Since the integral test is not allowed, you can use direct comparison test: Comparing your series to the harmonic series$\sum_{k=1}^{\infty}\frac{1}{k}$, ... View answer Accepted answer 1 votes I think the following works: $$\sum_{k=2}^{\lfloor\frac{n}{3}\rfloor} \sum_{j=1}^{min(k-1,n-3k-1)} \lceil \frac{n-(3 k+j)}{2} \rceil$$ Explanation The outer sum follows from your$3 \times k, ... View answer 1 votes This questions seems to be very similar to graph theory problems and as @MisraLavrov has showen, its solvable with methods from graph theory. One example of those methods is the concept of vertex-... View answer 0 votes \begin{align} \frac{2^{k+1}+(-1)^k}{3 \times 2^k}+\left(-\frac{1}{2}\right)^{k+1} &=\frac{2 \times 2^{k+1}+2 \times(-1)^k}{3 \times 2^{k+1}}+\frac{3 \times (-1)^{k+1}}{3 \times 2^{k+1}} \\ &=\... View answer 0 votesF(x,y)=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( \frac{x+3}{x}\...

Write $n!=2 \times \prod_{k=2}^{n-1}(1+k)$, then expand the product to get a subset of the divisors of $\frac{1}{2}n!$. Now you just need to multiply each term by $2$. Since you want one summand to be ...

Obviously the case $a=0$ holds. Since $$1-a=1-\frac{0.1a}{1-0.9a} \Rightarrow 1=\frac{0.1}{1-0.9a} \Rightarrow 1-0.9a=0.1a \Rightarrow 1=a,$$ your equations hold iff $a \in \{0,1\}$.

A very natural solution is to weight the values. For example take $f(x)=1 \times$ positive $+ 0 \times$ neutral $+ (-1) \times$ negative. It's easy to show that $f(x) \in [-1,1]$ and why you can ...

Sometimes there is only one of max and min. Take for example $f(x)=\frac{1}{x^2+1}$, which answers your question about "and" / "or". Also your argumentation does not work: Every ...

Hint: Write it as a product and use $1+\frac{1}{101-n}=\frac{102-n}{101-n}$, then you can apply the telescoping technique.
As mentioned in the comments, you can use a recurrence, which needs only $\frac{1}{2},+$ and $\times$: Since $u_n=\frac{1}{2}+\frac{1}{2^n} \Rightarrow \frac{1}{2^n}=u_n-\frac{1}{2}$, we have $u_{n+1}... View answer 0 votes In the following links there are many information about the function$sinc(x)$and it's values: Definition of Sinc function (another question with complete answers) https://en.m.wikipedia.org/wiki/... View answer 0 votes Not a answer, but important to know: As Qiaochu Yuan mentioned, the Gamma function gives your equality. But in the usual definition of the factorial it's undefined. That means the Gamma function is ... View answer 0 votes The argumentation in you're answer seems to be correct, although I haven't calculated it all. You should add the possibility, that P only "touches" the x-axis in$x^*\$. Consider for example ...