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Consider the cusp form $\Delta$ of weight 12 defined as: $$\Delta=2^{-6}3^{-3}(E_2^3-E_3^2)$$ where $E_2$ and $E_3$ are the normalised Eisenstein series that are modular forms of weight 4 and 6 respectively and are given as: $$E_2(z)=1+240\sum\limits_{n=1}^{\infty}\sigma_3(n)e^{2\pi inz}$$ $$E_3(z)=1-504\sum\limits_{n=1}^{\infty}\sigma_5(n)e^{2\pi inz}.$$ Here $\sigma_3(n)$ denotes the sum of the 3rd powers of the divisors of $n$, whereas $\sigma_5(n)$ denotes the sum of the 5th powers.

It is a standard result that $\Delta$ has integer Fourier coefficients. I wish to prove that that $1/\Delta$ has integer Fourier coefficients. I have encountered proofs which use this fact without proving it. Is it obvious?

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    Obvious enough that it's also used implicitly [here](http://math.stackexchange.com/a/532179/75794). – 2017-02-08
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    [Relevant topic](https://math.stackexchange.com/questions/723549/) – 2017-02-08

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The standard geometric series computation shows that $$ \dfrac{1}{1-X} = 1 + X + X^2 + \cdots.$$ Thus $$ \dfrac{1}{q(1-X)} = \dfrac{1}{q} + \dfrac{X}{q} + \dfrac{X^2}{q} + \cdots.$$ Now apply this with $$ X = - \sum_{n = 2} ^{\infty} \tau(n) q^{n-1},$$ to compute $1/\Delta$.

Since $X$ has integral $q$-expansion coefficients, so do all its powers, and hence so does $1/\Delta.$