think this statement is reasonable, if is f(x) is continuous and $f'(x)\leq 1/x$ for all postive x, then $\sum_{k=1}^{n} f(k)$ is assymptotic to $f(n)n$
Let f(k) be any integer function satisfying atleast, f(ab)=f(a)+f(b), when gcd(a,b)=1, if $$\sum_{n=1}^\infty \frac{f(k^s}{k^s}$$ converges for all s>1, then $$\sum_{k=1}^n f(k)=\sum_{p\leq n} f(p)[\frac{n}{p}]+o(n)
fuck sake
aaa
Let f(k) be any integer function satisfying atleast, f(ab)=f(a)+f(b), when gcd(a,b)=1, if $$\sum_{k} \frac{f(k^s)}{k^s}$$ converges for all s>1, then $$\sum_{k=1}^n f(k)=\sum_{p\leq n} f(p)[\frac{n}{p}]+o(n)$$
This Dirichlet series: $\sum _{n=0}^{\infty } \left(-\frac{1}{6 n+2}-\frac{2}{6 n+3}-\frac{1}{6 n+4}+\frac{1}{6 n+5}+\frac{2}{6 n+6}+\frac{1}{6 n+1}\right)$ converges to zero.
I have tried to modify it to make it reveal something about the Riemann zeta zeros but the set of tools in my toolbox is limited and I have not gotten far.
This is another one: $\sum _{n=0}^{\infty } \left(-\frac{1}{10 n+2}+\frac{1}{10 n+3}-\frac{1}{10 n+4}-\frac{4}{10 n+5}-\frac{1}{10 n+6}+\frac{1}{10 n+7}-\frac{1}{10 n+8}+\frac{1}{10 n+9}+\frac{4}{10 n+10}+\frac{1}{10 n+1}\right)$ that also converges to zero.
user19161
1:05 PM
@MatsGranvik I hope you solve the Riemann Hypothesis soon! =)
There is one in between: $\sum_{k=0}^\infty\left(\frac1{8n+1}-\frac1{8n+2}+\frac1{8n+3}-\frac5{8n+4}+\frac1{8n+5}-\frac1{8n+6}+\frac1{8n+7}+\frac3{8n+8}\right)$
Ok, I am not sure I understand how. The ones I mentioned are from the matrix a(GCD(n,k)) where "a" is the Dirichlet inverse of the Euler totient function. http://oeis.org/A191898
@MatsGranvik The first line is the alternating harmonic series written in one way, the second line is also the alternating harmonic series written in another way.
@MatsGranvik so you get $\log(2)-\log(2)=0$
@MatsGranvik The one for 8 could even get trickier...
@DumbCow One can also write the condition as $f(t+k)-f(t)=0$ so that $$f(t+nk)-f(t)=f(t+nk)+\sum_{i=1}^{n-1}(-f(t+ik)+f(t+ik))-f(t)$$ $$=\sum_{i=0}^{n-1}(f(t+(n+1)k)-f(t+nk))=0.$$
Robjohn became good at mathematics like most people who are good at mathematics have become good at mathematics: work hard. There are no shortcuts 8-).
@HenningMakholm It is pretty strange. Then you complain to the guy at the counter and they say 'don't complain to me! call this number <...> 25 ct/minute'
@PeterTamaroff I believe so - and I think none of them would be accepted today as complete - but then nobody had developed the necessary topological machinery in his day
I eco what a friend told me once: He said that when he learnt some songs he loved in the guitar, they lost their mystique. I feel somehow something similar here: when I see the proofs of some theorems, they seem to lose their mystique, their flavour.
@PeterTamaroff I get the same feeling sometimes, in a way - reading Gauss in D.A. - his arguments seem fresher in some way that I find hard to define - although his arguments can be a long-winded by modern standards.
for instance he didn't have the concept of a group - so he proved almost the same thing 3-4 times in D.A. when a modern book would just say "follows from Lagrange's theorem", I think
@PeterTamaroff I think that post was left vacant at the last ICM. It was felt that the various proofs of the FTA are now mature enough not to need constant caretaking.
