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

$\sum_{n=1}^\infty a_n<\infty$ if and only if $\sum_{n=1}^\infty \frac{a_n}{1+a_n}<\infty$

For $a_n$ positive sequence. I think I can prove one direction, but not both. Answer Since $\frac{a_n}{1+a_n}<a_n$, thus $\sum_{n=1}^\infty a_n<\infty$ implies $\sum_{n=1}^\infty \frac{a_n}{1+a_n}<\infty$. On the other hand, let $\sum_{n=1}^\infty \frac{a_n}{1+a_n}<\infty$. Then $\lim_{n\to\infty}\frac{a_n}{1+a_n}=0$ and hence $\lim_{n\to\infty}a_n=0$. So there is $N>0$ such that $a_n<1$ when $n\ge N$. From this, we have $$ \frac{1}{2}a_n\le\frac{a_n}{1+a_n}, \text{ for }n\ge N … Read more

Does fn(x)=cosn(x)(1−cosn(x))f_n(x)=\cos^n(x)(1-\cos^n(x)) converge uniformly for xx in [π/4 , π/2][π/4 , π/2]?

Does f_n(x)=\cos^n(x)(1-\cos^n(x)) converge uniformly for x in [π/4 , π/2]? Its clear to see that the point-wise convergence is to 0. By finding the derivative I obtained that the maximum of f_n is when \cos^n(x)=1/2 and that \sup |f_n(x)-0|= 1/2\cdot(1-1/2)=1/4 which would indicate that the function does not converge uniformly, however I’m not sure the … Read more

Is convergence in probability sometimes equivalent to almost sure convergence?

I was reading on sufficient and necessary conditions for the strong law of large numbers on this encyclopedia of math page, and I came across the following curious passage: The existence of such examples is not at all obvious at first sight. The reason is that even though, in general, convergence in probability is weaker … Read more

Uniqueness of a Limit epsilon divided by 2?

I have been reading about this theorem in a book called ‘Calculus: Basic Concepts for High-schools’, it is a very good book (so far) and I can highly recommend it. Well the author goes on to prove that: ‘A convergent sequence has only one limit’, which is quite intuitive in geometrical sense.. However in the … Read more

Task with convergence: ∫+∞0earctanxx2dx\int_{0}^{+\infty }\frac{e^{\arctan x}}{x^{2}}dx and ∫+∞0earctanx1+x2dx\int_{0}^{+\infty }\frac{e^{\arctan x}}{1+x^{2}}dx

a) \int_{0}^{+\infty }\frac{e^{\arctan x}}{x^{2}}dx b) \int_{0}^{+\infty }\frac{e^{\arctan x}}{1 + x^{2}}dx I know that is same , but I need to first check convergence which I think that when I put infinity e^{\frac{\pi}{2}} and then that get out and I have \int_{0}^{+\infty }\frac{1}{x^{2}}dx that convergent because p>1 And now I need to calculate it. And there … Read more

If ∑∞n=1an\sum_{n=1}^\infty a_n converges, prove that lim\lim_{n\to \infty} (1/n) \sum_{k=1}^n ka_k = 0.

the series a_n is any arbitrary converging series. My thought process was that the 1/n will definitely go to zero as n approaches infinity; however, the series k*a_k seems to approach infinity at the same rate. This confuses me because if both the numerator and denominator approach infinity, I think that the overall equation will … Read more

Does series with factorials converge/diverge: $\sum\limits_{n=1}^\infty \frac{4^n n!n!}{(2n)!}$?

$$\sum_{n=1}^\infty {{4^n n!n!}\over{(2n)!}}$$ I tried the ratio test but got that the limit is equal to 1, this tells me nothing of whether the series diverges or converges. if I didn’t make any errors when doing the ratio test, it may diverge, but I need help proving that. Is there any other test I could … 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