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In Regarding the $\sigma(n)$ function, I say in this site Mathematics Stack Exchange was stated in one of the answer, if I understand well, the following

Claim (Fischer). Let $\sigma(n)=\sum_{d\mid n}d$ the sum of divisor function, then for any positive integer $n\geq 1$ one has $$\frac{\sigma(n^2)}{n\sigma(n)}< \frac{\zeta(2)}{\zeta(4)}=\frac{15}{\pi^2}.$$ And it is possible improve this bound.

Thus, if I understand well Lagarias claim in his paper, I say the last of his comments about Kaneko's statement (currently I understand the proof of Lagarias lemmas, but not all paper his paper), page 542 of Lagarias, An Elementary Problem Equivalent to the Riemann Hypothesis American Mathematical MONTHLY June-July 2002, I can combine previous claims to state that for $n$ sufficiently large (around $n>60$) $$\sigma(n^2)< \frac{15}{\pi^2}ne^{H_n}\log H_n,$$ where we are denoting by $$H_k=1+\frac{1}{2}+\ldots+\frac{1}{k}$$ the kth harmonic number (that is this well arithmetic function), thus one has that for an integer $n\geq 1$ one can compute $H_{n^2}$ as definition to be $$H_{n^2}=\sum_{k=1}^{n^2}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n^2}.$$ My question is

Question. Can you provide us an upper bound for the difference (I am considering that it is a positive difference, I don't know how prove it) $$\frac{15}{\pi^2}ne^{H_n}\log(H_n)-e^{H_{n^2}}\log(H_{n^2})$$ for large positive integers $n$? I say calculations that I can learn and study, thus it isn't neccesary the best bound or work with the best Fischer's claim. Thanks in advance.

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  • Also I will try understand well cited question/answer in this site MSE. I hope that previous myself question has mathematical sense; also I am interesting in if it is possible combine the property that the sum of divisor function is multiplicative with calculations involving Robin, Lagarias or Kaneko equivalences. But it is a different question. –  Aug 09 '16 at 17:39
  • Please, all users are welcome to explore my ideas and exploit them, if there were some potentially useful. I hope that those aren't the worse, I say myself ideas/calculations, because currently I have no good abilities in mathematics, but I believe in effort and always I try that my point of departure is solid, that is the cause of I used the best references, because are the best advantage. –  Aug 09 '16 at 19:26

2 Answers2

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Since $\frac{1}{x}$ is decreasing, we have $\int_n^{n+1} \frac{1}{x}dx \leq \frac{1}{n} \leq \int_{n-1}^n \frac{1}{x}dx$, so: $$ \log(n+1) < \sum_{k=1}^n \frac{1}{k}=H_n < \log(n)+1 $$ Thus since $\log$ and $\exp$ are increasing: $$ \frac{15}{\pi^2}n(n+1)\log (\log(n+1))\leq \frac{15}{\pi^2}ne^{H_n} \log(H_n) \leq \frac{15}{\pi^2}n^2 e \log(\log(n)+1) $$ $$ (n^2+1) \log(\log(n^2+1))\leq e^{H_{n^2}} \log(H_{n^2}) \leq en^2\log(\log(n^2)+1)) $$ Taking difference we have tight upper and lower bounds? Not sure if this was what you're looking for?

user356869
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  • I did the calculations and the first was $1+\log(n+1)-log 2<H_n<1+\log n-log 2$, that is thus your first claim, the other calculations also I understand well. Thus I accept your answer, truly this kind of calculations should be able do myseflt. Very thanks much and thanks for the patience of all users. –  Aug 09 '16 at 18:35
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In the answer to this MSE question, Will Jagy hints that an upper bound for $\sigma(n^2)/\left(n\sigma(n)\right)$ is given by $$\dfrac{\sigma(n^2)}{n\sigma(n)} < \dfrac{\zeta(2)}{\zeta(3)} \approx 1.3684327776\ldots.$$

This improves on Fischer's upper bound of $$\dfrac{\zeta(2)}{\zeta(4)} \approx 1.519817754635\ldots.$$

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Jose Arnaldo Bebita Dris
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  • Very thanks much then to you for this nice information and Will Jagy. I see now those hints. –  Aug 12 '16 at 07:14