# Bound on an expression involving J-function coefficients

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).

• Have you tried looking for an interpretation of $(m + 1) c_m - \sum(\dots)$ as the $m$-th coefficient of something? Jan 24, 2019 at 21:19

Let us define $$j=J+744=\frac{1728E_4^3}{E_4^3-E_6^2}$$. Then, as noted by OP, the question is equivalent to positivity of coefficients of $$f=q\frac{d}{dq}J+E_2(J+24)=q\frac{d}{dq}j+E_2j-720E_2.$$ Next, by this answer, we have $$q\frac{d}{dq}j=-\frac{E_6}{E_4}j,$$ therefore $$f=j\left(E_2-\frac{E_6}{E_4}\right)-720E_2.$$ On the other hand, by Ramanujan identities (see here, for example), we have $$q\frac{d}{dq}E_4=\frac{E_2E_4-E_6}{3},$$ so $$f=3j\frac{q\frac{d}{dq}E_4}{E_4}-720E_2=5184\frac{E_4^2q\frac{d}{dq}E_4}{E_4^3-E_6^2}-720E_2.$$ As coefficients of $$E_4$$ are all non-negative, so are coefficients of its derivative. Also, $$\frac{1}{E_4^3-E_6^2}=\frac{1}{\Delta}=q^{-1}\prod_{n=1}^{+\infty}(1-q^n)^{-24}$$ has positive coefficients. Due to the fact that coefficients of $$E_2$$ (apart from the $$q^0$$ term) are non-positive, we obtain the desired inequality.

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 $$(m+1)e^{4\pi\sqrt{m}}$$.

• 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 $m-n=o(m)$?
– Ella
Jan 27, 2019 at 22:09
• @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. Jan 27, 2019 at 22:29
• @Ella if they are divisor sigmas times 24, the bounds in OP should be probably also multiplied by 24? Jan 28, 2019 at 7:36
• @Fedor Petrov Right, my bad! I'll edit the question.
– Ella
Jan 28, 2019 at 20:17