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I would like to show that $$(m+1)\,c_m- \sum_{n=1}^{m-1} c_n \,\sigma_{m-n} > 0$$ for all $m$, where $c_i$ is the $i$th coefficient of Klein's $J$-function $$J(q)= \frac{1728 \;E_4^3(q) }{E_4^3(q)-E_6^2(q)}-744$$ (in terms of Eisenstein series $E_4$ and $E_6$) and $\sigma_i$ is minus the $i$th coefficient of the second order Eisenstein series $E_2$. Both are positive numbers, bounded respectively by $$c_n \leqslant \frac{e^{4\pi\sqrt{n}}}{\sqrt{2}\;n^{3/4}}$$ $$c_n \geqslant \frac{e^{4\pi\sqrt{n}}}{\sqrt{2}\;n^{3/4}}\, K$$ $$\sigma_n < 24 \left[e^{\gamma}\,n\,\text{ln}(\text{ln}(n))+\frac{0.6483\, n}{\text{ln}(\text{ln}(n))}\right]\,\qquad (n\geqslant 3)$$ where $\gamma$ is Euler's constant ($\sim 0.6$) and $K\sim 0.98$. I could also use the cruder bound $$c_n \leqslant e^{4\pi \sqrt{n+1}}\,.$$ I've tried bounding the quotient of the two terms, Taylor-expanding and taking a limit to show that it goes to zero, but that's only valid for small values of $n$ and I'm interested in all kinds of values. Numerical evidence suggests this quotient goes to something like 1, but I don't know how to show that, if I must refrain from Taylor-expanding. (Also, replacing the sum by $(m-1)c_{m-1}\sigma_1$ is much too brutal).

Any ideas are welcome, and if there's any interesting information on the coefficients that I'm not aware of, I would appreciate also! For example, knowledge of the relations between successive coefficients might enable one to make a recursive type of argument, but I haven't found any such thing.

Edit: To be more complete and answer David Loeffler's comment, this expression appears in the modular form of weight 2 $$q\partial_q\,J(q)+E_2\left(J(q)+24\right)=\sum_{m=1}^{\infty}F(m)\,q^m$$ with $F(m)=(m+1)\,c_m- \sum_{n=1}^{m-1} c_n \,\sigma_{m-n}-\sigma_{m+1}-\sigma_1\sigma_m$ (I'm assuming the sigma terms are negligible).

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    $\begingroup$ Have you tried looking for an interpretation of $(m + 1) c_m - \sum(\dots)$ as the $m$-th coefficient of something? $\endgroup$ – David Loeffler Jan 24 at 21:19
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UPDATE Since I thought $\sigma_n$ was just the sum of the divisors of $n$ and as @Ella pointed out I was off by a factor of exactly $24$ what is written below is only the (sketch) of the proof of the fact that $\sum_{n = 1}^{m -1}c_n\sigma_{m - n}$ is asymptotic to $(m+1)c_m$ and almost nothing towards the desired inequality.

First of all note that we are not interested in the big $m-n$ because for those $n$ we have $c_n$ much smaller than $c_m$ and even their sum multiplied by sum of all $\sigma_n$(which is $O(m^3)$ say) will not overcome even one $c_m$. In particular we may assume that $m-n = o(m)$. For this range ratio of $\sqrt{2}m^{3/4}$ and $\sqrt{2}(m-n)^{3/4}$ is $1 + o(1)$ so we can drop these factors too. Also before going to summation let's switch $n$ and $m-n$ for brevity.

After preparations in the previous paragraph we are left with the following sum

$$\sum_{n = 1}^m \sigma_{n}e^{4\pi \sqrt{m-n}}.$$

Let $x > 0$ be a fixed small number. Let's split our sum into blocks of length $x\sqrt{m}$. We have $4\pi \sqrt{m - kx\sqrt{m}} \approx 4\pi \sqrt{m} - 2\pi kx$ (at least for $k = o(\sqrt{m})$, but as we written before we are interested only in the much smaller $n$'s actually so this approximation can easily be made effective). Since $x$ is small on each interval value of $e^{4\pi \sqrt{m-n}}$ are essentially the same and we will replace them by $e^{4\pi \sqrt{m} - 2\pi kx}$ Thus after this approximation we get

$$\sum_{k = 0}^\infty \sum_{n = kx\sqrt{m}}^{(k+1)x\sqrt{m}}\sigma_n e^{4\pi \sqrt{m}-2\pi kx} = e^{4\pi \sqrt{m}}\sum_{k = 0}^\infty e^{-2\pi kx}\sum_{n = kx\sqrt{m}}^{(k+1)x\sqrt{m}}\sigma_n.$$

(sum over $k$'s actually not to infinity but for the upper bound it is irrelevant while for the lower bound any $k$ will appear for big enough $m$'s).

We now need the following well-known asymptotic for sum of $\sigma$'s:

$$\sum_{n \le t} \sigma(n) \asymp 2\pi^2t^2.$$

(that's exactly our saving over uniform bound $\sigma_n \le cn\ln (\ln (n))$ which will give us only bound $ct^2\ln(\ln (t))$). Using this asymptotic we get

$$ e^{4\pi \sqrt{m}} \sum_{k = 0}^\infty e^{-2\pi kx}2\pi^2x^2m(2k+1) = 2\pi^2e^{4\pi \sqrt{m}}mx^2 \sum_{k = 0}^\infty (2k+1)e^{-2\pi kx}.$$

This last sum can be calculated explicitly and we finally get

$$2\pi^2e^{4\pi \sqrt{m}}mx^2 \frac{e^{2\pi x}(e^{2\pi x} + 1)}{(e^{2\pi x} - 1)^2}.$$

Recall that $x > 0$ is some small number so let's calculate limit of this expression as $x\to 0$. We get

$$ e^{4\pi \sqrt{m}}m 2\pi^2 \frac{2}{(2\pi)^2} = e^{4\pi\sqrt{m}}m.$$

and this expression up to $1+o(1)$ is $(m+1)e^{4\pi\sqrt{m}}$.

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  • $\begingroup$ Thanks a lot for your answer. I agree with everything, except for one tiny detail: the $\sigma_n$ in my expression is the $n$th coefficient of $E_2$, so they are divisor sigmas times 24. Thus the final result is that $m<m+1$, which is maybe a bit tight regarding the assumption that $m-n=o(m)$? $\endgroup$ – Ella Jan 27 at 22:09
  • $\begingroup$ @Ella Ouch, in that case I proved nothing regarding your inequality, sorry. I doubt that such a crude analysis can be pushed as far as to prove what we want. Even if it could, we should be more accurate with throwing out $n^{3/4}$ and big $m - n$'s, give explicit error terms in sum of $\sigma$'s and most importantly we will need much more precise estimate for $c_n$ than just asymptotic. Anyway I'll edit my answer accordingly. $\endgroup$ – Aleksei Kulikov Jan 27 at 22:29
  • $\begingroup$ @Ella if they are divisor sigmas times 24, the bounds in OP should be probably also multiplied by 24? $\endgroup$ – Fedor Petrov Jan 28 at 7:36
  • $\begingroup$ @Fedor Petrov Right, my bad! I'll edit the question. $\endgroup$ – Ella Jan 28 at 20:17

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