Sum of Sums over Primes that Divide the Index

drewfstr314
Messages
20
Reaction score
0
I have seen double sums, but have come across a problem involving sums over primes. However, this sum is inside a second sum, and is taken over all primes that divide the outside index, like this:

\sum_{k=1}^{n} \sum_{p | k} \frac 1p

for p prime.

Is there any way to manipulate this? Any help would be appreciated.

Thanks!
 
Mathematics news on Phys.org
I think that's equivalent to \sum_{p=2}^{n} \frac{\left \lfloor n/p \right \rfloor}{p}, where the square brackets represent the floor function, and p runs through the primes less than or equal to n.

I don't know if that helps at all, and no doubt it can be simplified more so.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top