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By specialization of an inequality I can write $$2 \sum_{k=1}^{n-1} \frac{1}{d_{k}} \sum_{l=k+1}^{n} \frac{1}{d_{l}}\leq 2\frac{\sigma_0(n)-1}{\sigma_0(n)}\cdot \left( \frac{\sigma(n)}{n} \right)^2, $$ where $1=d_1<d_2\ldots<d_{\sigma_{0}(n)}=n$ are the divisors of a positive integer $n>1$, $\sigma_0(n)$ is the number of such divisors and $\sigma(n)$ is the sum of divisor function. On the other hand $$\sum_{k=1}^{n-1} \frac{1}{d_{k}} \sum_{l=k+1}^{n} \frac{1}{d_{l}}=2 \left( \frac{\sigma(n)}{n} \right)^2-\frac{\sigma_2(n)}{n},$$ where $\sigma_2(n)=\sum_{d\mid n}d^2$. Thus by comparison of such formulas we can deduce $$(\sigma(n))^2\leq \sigma_0(n)\sigma_2(n).$$ After I've tried a comparison with my computer, known the average orders of these functions and involving also a different arithmetics function to adjust a conjecture. A graph for $$\frac{\sigma_0(n)\sigma_2(n)}{(\sigma(n))^2H_n}, $$ where $H_n=1+1/2+\cdot+1/n$ is the nth harmonic number, is below

below

Question. Can you say if is there limit as $n\to\infty$? Can do you explain, a heuristic, why the erratic behaviour of the graph at first? Thanks in advance.

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  • $\begingroup$ The cited inequality is the same reference of my recent post. $\endgroup$
    – user243301
    Commented Mar 21, 2016 at 15:51
  • $\begingroup$ It would not be surprising if the ratio without the extra $H_n$ term were bounded, in which case you would have a limit of zero. Compare Hardy and Wright, $\varphi(n) \sigma(n) / n^2$ is between $6/ \pi^2 $ and $1.$ The main thing is to identify types of numbers where that ratio is especially large, and those where it is especially small. For primes, it (yours) is about $2.$ $\endgroup$
    – Will Jagy
    Commented Mar 21, 2016 at 18:45
  • $\begingroup$ Right, Theorem 329 on page 267 of the fifth edition. What reference do you mean? $\endgroup$
    – Will Jagy
    Commented Mar 21, 2016 at 18:49
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    $\begingroup$ Very thanks much, @WillJagy sorry by the comments since I don't understand english well. Very thanks much for your notes, i try think about them, and feel free to add an answer $\endgroup$
    – user243301
    Commented Mar 21, 2016 at 19:45

1 Answer 1

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Your inequality $$(\sigma(n))^2\leq \sigma_0(n)\sigma_2(n)$$ is the Cauchy-Schwarz inequality in $\mathbb R^{\sigma_0(n)}.$ One of the vectors used in Cauchy Schwarz has all entries equal to $1,$ the other has all the divisors of $n$ as the entries.

Spanish Primorial

I tried it for the primorials, it would appear that there is no upper bound, even when dividing by your $H_n.$

 m 2 =  2 ratio  1.11111 ratio / log m 1.60299
 m 6 =  2 3 ratio  1.38889 ratio / log m 0.775154
 m 30 =  2 3 5 ratio  2.00617 ratio / log m 0.589843
 m 210 =  2 3 5 7 ratio  3.13465 ratio / log m 0.586232
 m 2310 =  2 3 5 7 11 ratio  5.31148 ratio / log m 0.685795
 m 30030 =  2 3 5 7 11 13 ratio  9.2138 ratio / log m 0.89368
 m 510510 =  2 3 5 7 11 13 17 ratio  16.4938 ratio / log m 1.25494
 m 9699690 =  2 3 5 7 11 13 17 19 ratio  29.8538 ratio / log m 1.8557
 m 223092870 =  2 3 5 7 11 13 17 19 23 ratio  54.9393 ratio / log m 2.85799
 m 6469693230 =  2 3 5 7 11 13 17 19 23 29 ratio  102.798 ratio / log m 4.5505
 m 200560490130 =  2 3 5 7 11 13 17 19 23 29 31 ratio  193.147 ratio / log m 7.42177
 m 7420738134810 =  2 3 5 7 11 13 17 19 23 29 31 37 ratio  366.498 ratio / log m 12.3669
 m 304250263527210 =  2 3 5 7 11 13 17 19 23 29 31 37 41 ratio  698.922 ratio / log m 20.9579
 m 13082761331670030 =  2 3 5 7 11 13 17 19 23 29 31 37 41 43 ratio  1335.75 ratio / log m 35.9943
 m 614889782588491410 =  2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 ratio  2562.51 ratio / log m 62.5609
 m 32589158477190044730 =  2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 ratio  4938.71 ratio / log m 109.919
 m 1922760350154212639070 =  2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 ratio  9553.67 ratio / log m 194.941
 m 117288381359406970983270 =  2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 ratio  18500.9 ratio / log m 348.292

Here is the same computer run, but i have saved space by not printing m or its factorization, just the largest prime dividing m.

 biggest prime factor 2 ratio  1.11111 ratio / log m 1.60299
 biggest prime factor 3 ratio  1.38889 ratio / log m 0.775154
 biggest prime factor 5 ratio  2.00617 ratio / log m 0.589843
 biggest prime factor 7 ratio  3.13465 ratio / log m 0.586232
 biggest prime factor 11 ratio  5.31148 ratio / log m 0.685795
 biggest prime factor 13 ratio  9.2138 ratio / log m 0.89368
 biggest prime factor 17 ratio  16.4938 ratio / log m 1.25494
 biggest prime factor 19 ratio  29.8538 ratio / log m 1.8557
 biggest prime factor 23 ratio  54.9393 ratio / log m 2.85799
 biggest prime factor 29 ratio  102.798 ratio / log m 4.5505
 biggest prime factor 31 ratio  193.147 ratio / log m 7.42177
 biggest prime factor 37 ratio  366.498 ratio / log m 12.3669
 biggest prime factor 41 ratio  698.922 ratio / log m 20.9579
 biggest prime factor 43 ratio  1335.75 ratio / log m 35.9943
 biggest prime factor 47 ratio  2562.51 ratio / log m 62.5609
 biggest prime factor 53 ratio  4938.71 ratio / log m 109.919
 biggest prime factor 59 ratio  9553.67 ratio / log m 194.941
 biggest prime factor 61 ratio  18500.9 ratio / log m 348.292
 biggest prime factor 67 ratio  35929.5 ratio / log m 626.784
 biggest prime factor 71 ratio  69890.7 ratio / log m 1134.84
 biggest prime factor 73 ratio  136055 ratio / log m 2065.29
 biggest prime factor 79 ratio  265392 ratio / log m 3778.02

edit: I understand now: your function ( without the $H_n$) is multiplicative in the number theory sense. This means that multiplying by a new prime $p$ multiplies the "ratio" above by the function applied to $p$ itself, which is approximately $2.$ So the number I call "ratio" above is roughly doubling each time. Let $$ f(n) =\frac{\sigma_0(n)\sigma_2(n)}{(\sigma(n))^2}. $$

 biggest prime factor 2 ratio  1.11111 cumulative product 1.11111
 biggest prime factor 3 ratio  1.25 cumulative product 1.38889
 biggest prime factor 5 ratio  1.44444 cumulative product 2.00617
 biggest prime factor 7 ratio  1.5625 cumulative product 3.13465
 biggest prime factor 11 ratio  1.69444 cumulative product 5.31148
 biggest prime factor 13 ratio  1.73469 cumulative product 9.2138
 biggest prime factor 17 ratio  1.79012 cumulative product 16.4938
 biggest prime factor 19 ratio  1.81 cumulative product 29.8538
 biggest prime factor 23 ratio  1.84028 cumulative product 54.9393
 biggest prime factor 29 ratio  1.87111 cumulative product 102.798
 biggest prime factor 31 ratio  1.87891 cumulative product 193.147
 biggest prime factor 37 ratio  1.89751 cumulative product 366.498
 biggest prime factor 41 ratio  1.90703 cumulative product 698.922
 biggest prime factor 43 ratio  1.91116 cumulative product 1335.75
 biggest prime factor 47 ratio  1.9184 cumulative product 2562.51
 biggest prime factor 53 ratio  1.9273 cumulative product 4938.71
 biggest prime factor 59 ratio  1.93444 cumulative product 9553.67
 biggest prime factor 61 ratio  1.93652 cumulative product 18500.9
 biggest prime factor 67 ratio  1.94204 cumulative product 35929.5
 biggest prime factor 71 ratio  1.94522 cumulative product 69890.7
 biggest prime factor 73 ratio  1.94668 cumulative product 136055
 biggest prime factor 79 ratio  1.95062 cumulative product 265392
 biggest prime factor 83 ratio  1.95295 cumulative product 518296
 biggest prime factor 89 ratio  1.95605 cumulative product 1.01381e+06
 biggest prime factor 97 ratio  1.9596 cumulative product 1.98667e+06

Alright: from Merten's Theorem, if prime number $k$ is called $p,$ so that $p = p_k,$ the "cumulative product" is approximately a constant times $$ \frac{2^k}{\log^2 p} $$

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  • $\begingroup$ Very thanks much @WillJagy for this answer and your approach, and references, I will read your notes. I understand that your have used the second theorem due to Mertens. Isn't? $\endgroup$
    – user243301
    Commented Mar 22, 2016 at 6:50

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