## Concerning the sequence ∑p≤n,p⌊lognlogp⌋/n \sum_{p \le n , p} \big\lfloor\frac{\log n}{\log p}\bigr\rfloor/n where p is prime

The sequence (∑p≤n,pisprime⏟⌊lognlogp⌋n)∞n=1 is bounded , my question is what is the lim sup of this sequence ? Is the sequence convergent ? If it does , what is the limit ? Answer We have1n∑p≤n⌊log(n)log(p)⌋=1n∑p≤n|{m∈N+:pm≤n}|=1n∑pm≤n1=π(n)n+O(log(n)√n)=O(1log(n)+log(n)√n)=O(log(n)√n). where π(n)=∑p≤n1. Expanding a little bit, note that 1)|{m∈N+:pm≤x}|=|{m∈N+:mlog(p)≤log(x)}|=⌊log(x)log(p)⌋ 2)We have the bound |{p:pm≤x,m≠1}|=O(√xlog(x)) 3)By Prime Number Theorem π(x)=(xlog(x)). AttributionSource : Link … Read more

## Definition of Strong Convergence in LpL^p

Is strong convergence in Lp, ie fistrongly⟶, just ||fi−f||Lp→0. If so why dont we just call it convergence? Answer Yes your definition is correct. Usually people do simply say convergence, but if you want to really emphasise that you do not mean another form of convergence e.g. weak convergence or almost everywhere convergence, you can … Read more

## Convergence of \int_1^\infty \frac{dt}{t^4+t^2+1}\int_1^\infty \frac{dt}{t^4+t^2+1}

I have to calculate \displaystyle\int_1^\infty \frac{dt}{t^4+t^2+1} and i have as a hint: Divide by t^2+t+1 Then I get that (t^2+t+1)(t^2-t+1)= t^4+t^2+1 I could do this by partial fractions but it would be tedious… Is there any other way that I am not seeing? Answer HINT: Notice, \int_{1}^{\infty}\frac{dt}{t^4+t^2+1}=\lim_{z\to \infty}\int_{1}^{z}\frac{dt}{t^4+t^2+1} =\lim_{z\to \infty}\int_{1}^{z}\frac{\left(\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+1} =\frac{1}{2}\lim_{z\to \infty}\int_{1}^{z}\frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}+1}\ dt =\frac{1}{2}\lim_{z\to \infty}\left(\int_{1}^{z}\frac{\left(1+\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+1}\ … Read more