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1h
comment On modulus of powers
@Wojowu: You cannot take $c=b+a$, because of the condition for $i=1$.
2h
revised What is the best way to learn about Modular Forms?
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1d
revised Elementary congruences and L-functions
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1d
awarded  Nice Answer
1d
revised Elementary congruences and L-functions
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1d
comment Elementary congruences and L-functions
I liked your formula for $(T/N)$ a lot, and I provided a proof below.
1d
revised Elementary congruences and L-functions
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1d
revised Elementary congruences and L-functions
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1d
revised Elementary congruences and L-functions
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1d
answered Elementary congruences and L-functions
2d
revised Asymptotics for the number of abelian groups of order at most $x.$
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2d
comment Use of infinitude of primes in the Green-Tao theorem
I think this proof is not circular. On the other hand, as I recall, Green and Tao do use basic facts like $\zeta(s)$ has a pole at $s=1$ which implies readily that there are infinitely many primes.
2d
comment A decreasing sequence involving the divisor function?
@PaceNielsen: I see, I did not look at that preprint for lack of time. I just verified on the back of an envelope earlier that the limit is $\frac{6}{\pi^2}e^\gamma$, from which the inequality follows for large $k$. This was the original question. I understand now that it holds for all $k$. Thanks for pointing this out.
2d
comment A decreasing sequence involving the divisor function?
@PaceNielsen: See my comment below my answer (a response to @favoured). The RH is not needed here, because $N_k$ is a very specific sequence for which the limit can be calculated.
2d
comment A decreasing sequence involving the divisor function?
@favoured: Yes, because the fraction tends to $6/\pi^2$ times the right hand side.
Feb
11
revised A decreasing sequence involving the divisor function?
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Feb
11
answered A decreasing sequence involving the divisor function?
Feb
11
revised A decreasing sequence involving the divisor function?
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Feb
11
comment A decreasing sequence involving the divisor function?
You misquote Gronwall's theorem which is about $\limsup_{N\to\infty}\frac{\sigma(N)}{N\log\log N}$ rather than about $\limsup_{k\to\infty}\frac{\sigma(N_k)}{N_k\log\log N_k}$. In fact it is easy to show that the latter equals $\frac{6}{\pi^2}e^\gamma=1.0827\dots$.
Feb
10
comment Analytic continuation of the double sum $\sum_{n,m\ge0}x^ny^mt^{nm}$
You don't need absolute convergence, locally uniform convergence is sufficient.