All Questions
13
questions
2
votes
0
answers
49
views
Weird computation on the (variant) divisor problem
I have a weird computations here about the (variation of the) divisor problems that involves the squarefull numbers. It is the problem to determine
$\displaystyle \sum_{a^2b^3\le x} 1,$ which is ...
- 1,033
1
vote
1
answer
183
views
How summation is changed in Analytic number theory
Consider this expression S(x, z) = $\sum_{n\leq x} \sum_{{d|n , d|P(z) } }\mu(d) $ . I don't understand the logic behind next step and get really confused on how summation is changed.
In next step ...
user775699
1
vote
1
answer
152
views
Does $\lim_{n \to \infty}\sum_{k=1}^n \left[\zeta\left(2k-1-\frac{1}{2n}\right) + \zeta(2k)\right]$ equal the Euler-Mascheroni constant?
Let $\zeta(s)$ be the Riemann zeta function and $\gamma$ be the Euler-Mascheroni constant. Is the following formula for the Euler-Mascheroni constant true?
$$
\lim_{n \to \infty}\sum_{k=1}^n \left[\...
- 20.8k
7
votes
1
answer
298
views
Are there infinitely many zeroes of $\sum_{r = 1}^{n-1} \mu(r)\gcd(n,r) $?
Let $\mu(n)$ be the Möbius function and $S(x)$ be the number of positive integers $n \le x$ such that
$$
\sum_{r = 1}^{n-1} \mu(r)\gcd(n,r) = 0
$$
My experimental data for $n \le 6 \times 10^5 $...
- 20.8k
14
votes
1
answer
640
views
Relationship between GCD, LCM and the Riemann Zeta function
Let $\zeta(s)$ be the Riemann zeta function. I observed that as for large $n$, as $s$ increased,
$$
\frac{1}{n}\sum_{k = 1}^n\sum_{i = 1}^{k} \bigg(\frac{\gcd(k,i)}{\text{lcm}(k,i)}\bigg)^s \approx \...
- 20.8k
2
votes
1
answer
108
views
$\sum_{p\leq n}\sum_{q\leq N}\sum_{n\leq N;p|n,q|n}1=^? \sum_{pq \leq N}\big( \frac N{pq}+O(1)\big)+\sum_{p \leq N}\big( \frac N{p}-\frac N{p^2}\big)$
I was studying Marius Overholt 'A course in Analytic Number Theory'. There in the section of "Normal order method". The proposition he is going to prove is
$Var[w]=O(loglog(N))$ where $w(n)$ is ...
user533344
0
votes
1
answer
133
views
Partial Harmonic Sum Equals $1/\Pi(1-1/p)$
Given $n\in \mathbb{N}$, I would like to know why (preferably with an algebraic or analytic argument)
$$\bigg(\Pi_{p\le n,\,p\, \text{prime}}(1-1/p)\bigg)\sum_{i}1/i=1,$$
where the sum is over all $i=...
- 3,918
4
votes
1
answer
637
views
Identity for the divisor function: $\tau(mn)=\sum\limits_{d\mid(m,n)}\mu(d) \tau(m/d)\tau(n/d)$
Let $\tau$ denote the classical divisor function and $\mu$ be the
Möbius function.
Then for each pair of integers $n,m$ we have
$$\tau(mn)=\sum_{d\mid(m,n)}\mu(d) \tau(m/d)\tau(n/d),$$
where the ...
- 1,061
2
votes
1
answer
121
views
Why do the first two terms of Euler's summation by parts formula not cancel each other out?
Euler's summation by parts formula states that:
$$
\sum_{y < n \leq x} f(n) = \int_y^x{f(t)dt} + \int_y^x(t - \lfloor t \rfloor)f'(t)dt +f(x)(\lfloor x \rfloor - x) -f(y)(\lfloor y \rfloor -y)$$ (...
- 347
0
votes
1
answer
188
views
partial summation $\displaystyle{\sum_{p\leq X}f(p)/p}$ via $\displaystyle{\sum_{p\leq X}f(p)\log p}$.
In some paper I saw that to show that some inequality of this type
$$
\sum_{p\leq X}\frac{f(p)}p>\frac{\log^2X}{C_1}
$$
it says the above inequality followes using by partial summation if we show
$$...
- 929
1
vote
2
answers
155
views
Can we prove summation formula for the first $n$ terms of natural numbers through calculus? [closed]
Can we prove summation formula for the first $n$ terms of natural numbers through calculus?
What about the summation of first $n$ numbers of the form $a^k$ and other summation formulas like sum of a ...
- 2,411
1
vote
1
answer
973
views
Alternative proofs that Dirichlet products are associative?
Is there alternative proof of the following fact:
Dirichlet product on arithmetic function is associative.
I'm looking for something different than that given in Dirichlet's product with ...
- 2,851
4
votes
3
answers
4k
views
How to prove $\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$
For every positive integer $d$, we let $\tau\left(d\right)$ be the number of positive divisors of $d$.
Prove that
\begin{align}
\sum_{d|n} \tau^3(d)
= \left(\sum_{d|n} \tau (d)\right)^2
\end{align}
...
- 2,879