Recent content by Bibubo

Let $\left\{ p_{1},p_{2},\dots,p_{h}\right\}$ a set of consecutive prime numbers. I want to show that, if $h$ is large enough, then doesn't exists a natural number $n$ such that $$n\equiv1\textrm{ mod }p_{i},\,\forall i=1,\dots,h.$$ I think is true but I have no idea how to prove it. Am I wrong?

Thank you so much for your observations! I will try to fix these errors.

I think I get a solution, but I'm not sure. Suppose for absurd that $$\left\{ p\, primes:\, p\mid\underset{i=1}{\overset{n}{\sum}}a_{i}^{k},\, k\in\mathbb{N}\right\} =\left\{ p_{1},\dots,p_{r}\right\}.$$Let $a_{1}+\dots+a_{n}=c$ and let $b_{i}$ the order of $c$ as a element of...

A friend of mine proposed to me this problem some days ago.

Let $a_{1},\dots,a_{n},\, n>2$ positive and distinct integer. Prove that the set of primes divisors of the numbers $$a_{1}^{k}+\dots+a_{n}^{k}$$with $k\in\mathbb{N}$ is infinite.

How can I solve (or evaluating) this integral $$\int_{5}^{\infty}\frac{1}{\log\left(t\right)t^{r}}dt$$ with $r\geq 2$?

You can write as you told. In fact you have $$\frac{\left(1+3^{r}\right)}{1+3^{2r}}^{2}=\frac{\left(3^{r}\left(1+\frac{1}{3^{r}}\right)\right)^{2}}{3^{2r}\left(1+\frac{1}{3^{2r}}\right)}=\frac{\left(1+\frac{1}{3^{r}}\right)^{2}}{1+\frac{1}{3^{2r}}}$$ but I haven't understood how your proof can...

Sorry but I don't see the point. You have something like $$\log\left(\frac{\left(1+3^{r}\right)^{2}}{1+3^{2r}}\right)=2\left(\underset{n\, odd}{\sum}\frac{1}{n}\,\frac{1}{3^{rn}}-2\underset{n\, even,\,4\nmid n}{\sum}\frac{1}{n}\,\frac{1}{3^{2rn}}\right)$$ because if $4\mid n$ we have a...

I've tried the Euler product way, but I haven't found a solution.

Prove that for $r>2$ we have $$\frac{\zeta\left(r\right)}{\zeta\left(2r\right)}<\left(1+\frac{1}{2^{r}}\right)\frac{\left(1+3^{r}\right)^{2}}{1+3^{2r}}.$$ I've tried to write Zeta as Euler product but I haven't solve it.

Grat job! I have only one question... if you put $x=1$ in your identity you obtain $$\underset{k=1}{\overset{n}{\sum}}\frac{\left(-1\right)^{k}k}{\left(n+k\right)\left(2k+1\right)\left(1+2k\right)}\,\dbinom{n}{k}\dbinom{n+k}{k}$$ and it isn't like the left side of (*) because there is a another...

Yes I see what you say... I've think to use in some way the Legendre shifted polynomials $$P_{N}\left(x\right)=\left(-1\right)^{N}\underset{i=0}{\overset{N}{\sum}}\dbinom{N}{i}\dbinom{N+i}{i}\left(-x\right)^{i}$$ because the sums are very similar, and because we have...

I have this sum $$\left(N+1\right)^{2}\underset{j=1}{\overset{N}{\sum}}\frac{\left(-1\right)^{j}}{2j+1}\dbinom{N}{j}\dbinom{N+j}{j-1}\underset{i=1}{\overset{N}{\sum}}\frac{\left(-1\right)^{i}}{\left(2i+1\right)\left(i+j\right)}\dbinom{N}{i}\dbinom{N+i}{i-1}$$ and numerical test indicates that is...