How to obtain an upper bound for $\sum_{d|n} d \sum_{m|d} m$?

Question

I would like to obtain an upper bound for $$ S(n) = \sum_{d|n} d \sum_{m|d} m. $$ I can prove $S(n) \ll d(n) n^2$ where $d(n)$ is the number of divisors of $n$. But in the paper I am reading it states without explanation $$ S(n) \ll n^3/\phi^2(n) $$ where $\phi$ is the Euler totient function. I would appreciate any explanation of this. Thank you!

Answer 1

just to register something, the exponent on $\phi(n)$ is wrong, the expression should be $n^3/\phi(n)$

One piece is theorem 329 in Hardy and Wright $$ \frac{6}{\pi^2} < \frac{\sigma(n) \phi(n)}{n^2} < 1 $$

Just something to do. The procedure, due to Ramanujan, for surprisingly large values of your function, is this: begin with a real $\delta > 0.$ For that value, define a number $$ N_\delta $$ that is the $n$ value that maximizes the ratio $$ \frac{\sum_{d|n} \, d \, \sigma(d) \;}{n^{2 + \delta}} $$

For all but a certain countable set of $\delta,$ there will be just one maximizer; further, since everything is multiplicative, we find the best exponent $k$ for a prime $p$ by letting $p^k$ maximise (among powers of prime $p$) the same ratio, $$ \frac{\sum_{d|p^k} \, d \, \sigma(d) \;}{p^{2k + k \delta}} $$ What you do is compare that ratio to the ratios with $k+1$ instead, or $k-1$ instead. The end of the process is that $k$ is the floor of some complicated expression in $p$ and $\delta. \; \; $ Oh, the set of primes worth using is finite, when $p$ is too large, taking $k=1$ is worse than $k=0.$ There is a fairly gentle introduction at NICOLAS. Also helpful would be Alaoglu and Erdos.

There may or may not be a closed form process for Ramanujan's method. In the meantime, here are the first few extremal numbers. We knoe it is working properly when each new line is the previous line multiplied by a single prime, either increase one of the existing exponents, or multiply by a new prime. The list below is comparable to the Superior Highly Composite Numbers or the Colossally Abundant Numbers. The first, decimal number on each line is $\delta.$ The next is the best integer, $N_\delta,$ and its factorization. Note the prime factors begin with $2$ and are consecutive, while the exponents in each line are non-increasing. In short, the number is a product of primorials. Finally, the "ratio" is $$ \frac{\sum_{d|n} \, d \, \sigma(d) \;}{n^{2 + \delta}} $$

0.5 ;  2 =  2  ratio  1.237436867076458
0.33 ;  6 =  2 3  ratio  1.399422908205721
0.2 ;  12 =  2^2 3  ratio  1.922262330287559
0.14 ;  24 =  2^3 3  ratio  2.24193487709251
0.1 ;  120 =  2^3 3 5  ratio  2.687547369136427
0.09 ;  360 =  2^3 3^2 5  ratio  2.837687177098467
0.075 ;  2520 =  2^3 3^2 5 7  ratio  3.116057606451912
0.07 ;  5040 =  2^4 3^2 5 7  ratio  3.241377998542384
0.039 ;  55440 =  2^4 3^2 5 7 11  ratio  4.226267503969329
0.034 ;  110880 =  2^5 3^2 5 7 11  ratio  4.465016977358566
0.031 ;  1441440 =  2^5 3^2 5 7 11 13  ratio  4.623701782662667
0.03 ;  4324320 =  2^5 3^3 5 7 11 13  ratio  4.69282633602653
0.024 ;  21621600 =  2^5 3^3 5^2 7 11 13  ratio  5.14645345148709
0.021 ;  367567200 =  2^5 3^3 5^2 7 11 13 17  ratio  5.418933455278657
0.018 ;  6983776800 =  2^5 3^3 5^2 7 11 13 17 19  ratio  5.754532366348787
0.017 ;  13967553600 =  2^6 3^3 5^2 7 11 13 17 19  ratio  5.886759009794376
0.014 ;  321253732800 =  2^6 3^3 5^2 7 11 13 17 19 23  ratio  6.317087431469826
0.0104 ;  9316358251200 =  2^6 3^3 5^2 7 11 13 17 19 23 29  ratio  6.949519234316012
0.0103 ;  65214507758400 =  2^6 3^3 5^2 7^2 11 13 17 19 23 29  ratio  6.971417630032335
0.0100 ;  195643523275200 =  2^6 3^4 5^2 7^2 11 13 17 19 23 29  ratio  7.038710026874346
0.0090 ;  6064949221531200 =  2^6 3^4 5^2 7^2 11 13 17 19 23 29 31  ratio  7.287660709217948
0.0080 ;  12129898443062400 =  2^7 3^4 5^2 7^2 11 13 17 19 23 29 31  ratio  7.559967353802027
0.0070 ;  448806242393308800 =  2^7 3^4 5^2 7^2 11 13 17 19 23 29 31 37  ratio  7.861709906083687
0.0065 ;  18401055938125660800 =  2^7 3^4 5^2 7^2 11 13 17 19 23 29 31 37 41  ratio  8.02744821972626
0.0060 ;  791245405339403414400 =  2^7 3^4 5^2 7^2 11 13 17 19 23 29 31 37 41 43  ratio  8.215288631708901
0.0050 ;  37188534050951960476800 =  2^7 3^4 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47  ratio  8.63965680876621

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Answer 2

I can show the following (modulo my usual mistakes):

$\sum_{m=1}^{\infty} \dfrac{S(m)}{m^k} =\zeta(k)\zeta^2(k-1)$ and $ \dfrac{\sigma^2(n)\phi(n)}{n} \lt S(n) \lt \sigma^2(n) $.

Here's how.

As Crostul suggests, if $S(n) = \sum_{d|n} d \sum_{m|d} m $ then

$\begin{array}\\ S(p^k) &= \sum_{d|p^k} d \sum_{m|d} m\\ &= \sum_{j=0}^k p^j \sum_{m|p^j} m\\ &= \sum_{j=0}^k p^j \sum_{i=0}^jp^i\\ &= \sum_{j=0}^k p^j \dfrac{p^{j+1}-1}{p-1}\\ &= \dfrac1{p-1}\sum_{j=0}^k p^j(p^{j+1}-1)\\ &= \dfrac1{p-1}\left(\sum_{j=0}^k p^jp^{j+1}-\sum_{j=0}^k p^j\right)\\ &= \dfrac1{p-1}\left(p\sum_{j=0}^k p^{2j}-\dfrac{p^{k+1}-1}{p-1}\right)\\ &= \dfrac1{p-1}\left(\dfrac{p(p^{2(k+1)}-1}{p^2-1}-\dfrac{p^{k+1}-1}{p-1}\right)\\ &= \dfrac1{(p-1)^2(p+1)}\left(p(p^{2(k+1)}-1)-(p+1)(p^{k+1}-1)\right)\\ &= \dfrac1{(p-1)^2(p+1)}\left(p^{2(k+1)+1}-p-(p^{k+2}+p^{k+1}-p-1)\right)\\ &= \dfrac1{(p-1)^2(p+1)}\left(p^{2k+3}-p^{k+2}-p^{k+1}+1)\right)\\ &= \dfrac1{(p-1)^2(p+1)}\left(p^{2k+3}-p^{k+1}-p^{k+2}+1)\right)\\ &= \dfrac1{(p-1)^2(p+1)}\left(p^{k+1}(p^{k+2}-1)-(p^{k+2}-1))\right)\\ &= \dfrac1{(p-1)^2(p+1)}\left((p^{k+1}-1)(p^{k+2}-1)\right)\\ &= \dfrac1{(p+1)}\dfrac{(p^{k+1}-1)(p^{k+2}-1)}{(p-1)^2}\\ &= \dfrac1{(p+1)}\sigma(p^k)\dfrac{(p^{k+2}-1)}{p-1}\\ &= \dfrac1{(p+1)}\sigma(p^k)\dfrac{(p^{k+2}-p+p-1)}{p-1}\\ &= \dfrac1{(p+1)}\sigma(p^k)\left(\dfrac{p^{k+2}-p}{p-1}+1\right)\\ &= \dfrac1{(p+1)}\sigma(p^k)\left(p\sigma(p^k)+1\right)\\ &= \dfrac1{(p+1)}\sigma(p^k)\left((p+1)\sigma(p^k)-\sigma(p^k)+1\right)\\ &= \dfrac1{(p+1)}\sigma(p^k)(p+1)\sigma(p^k)-\dfrac1{(p+1)}\sigma(p^k)(\sigma(p^k)-1)\\ &= \sigma^2(p^k)-\dfrac{\sigma(p^k)(\sigma(p^k)-1)}{(p+1)}\\ &= \sigma^2(p^k)\left(1-\dfrac{(1-1/\sigma(p^k)}{(p+1)}\right)\\ &< \sigma^2(p^k)\\ \text{so}\\ S(n) &\lt \sigma^2(n)\\ \text{and}\\ S(p^k) &> \sigma^2(p^k)\left(1-\dfrac{1}{(p+1)}\right)\\ &= \sigma^2(p^k)\left(\dfrac{p}{(p+1)}\right)\\ &> \sigma^2(p^k)\left(\dfrac{p-1}{p}\right)\\ &= \sigma^2(p^k)\left(\dfrac{p-1}{p}\right)\\ &= \sigma^2(p^k)\dfrac{\phi(p^k)}{p^k}\\ \text{so}\\ S(n) &\gt \dfrac{\sigma^2(n)\phi(n)}{n}\\ \text{also}\\ S(n) &= \sum_{d|n} d \sum_{m|d} m\\ &= \sum_{m|n}\sum_{d|\frac{n}{m}} d m\\ &= \sum_{m|n}m\sum_{d|\frac{n}{m}} d\\ &= \sum_{m|n}m\sigma(\frac{n}{m})\\ \text{so if}\\ s(k) &=\sum_{m=1}^{\infty} \dfrac{S(m)}{m^k},\\ u(k) &=\sum_{m=1}^{\infty} \dfrac{m}{m^k}\\ &=\zeta(k-1)\\ \text{and}\\ v(k) &=\sum_{m=1}^{\infty} \dfrac{\sigma(m)}{m^k}\\ &=\zeta(k)\zeta(k-1)\\ \text{then}\\ s(k) &=u(k)v(k)\\ &=\zeta(k)\zeta^2(k-1)\\ \end{array} $