Direct proof of the non-zeroness of an Eisenstein series Question: Can you show directly from its formula that $G_4(i)\neq0$?
Recall that the holomorphic Eisenstein series of weight $2k$ is defined by:
$$G_{2k}(\tau)= \sum_{(m,n)\in\mathbb{Z}^2\setminus (0,0)} \frac{1}{(m+\tau n)^{2k}}.$$
Motivation Exercise 6.6 of Silverman's "The Arithmetic of Elliptic Curves" asks to compute the special value:
$$j(i)= j(\mathbb{Z} \oplus i\mathbb{Z})=1728.$$
Where $j(\tau)=j(\mathbb{Z} \oplus \tau \mathbb{Z}) = 1728 \frac{g_2(\tau)^3}{g_2(\tau)^3-27g_3(\tau)^2}$ is the $j$-invariant function.
To prove it it is enough to show $\frac{1}{140}g_3(i):=G_6(i)=0$ and $\frac{1}{60}g_2(i):=G_4(i)\neq 0$. 
Using the relation $\frac{1}{(m+in)^6}=-\frac{1}{(n-im)^6}$ I could prove $G_6(i)=0$. Then I  can use the fact that $\Delta(\tau)=g_2(\tau)^3 -27g_3(\tau)^2$ is never zero to deduce $G_4(i)\neq 0$ and prove the exercise. 
Before coming up with this stupid observation I spent quite some time in trying to prove $G_4(i)\neq0$ directly from the formula, but I didn't succeed. 
Thank you very much!
 A: Nice question.
There is a theorem of Hurwitz that says that 
$$G_4(i) = \frac{16}{15} \left(\int_0^1 \frac{dt}{\sqrt{1-t^4}}\right)^4 = \frac{16}{15}\left(\frac{\sqrt{\pi}\cdot \Gamma(5/4)}{\Gamma(3/4)}\right)^4 = 3.151212\ldots.$$
so it is positive. But you don't need the exact value, you just need to know that $G_4(i)\neq 0$. For that, rewrite $G_{2k}$ as follows:
$$G_{2k}(\tau)=\sum_{(m,n)\neq (0,0)} \frac{1}{(m+n\tau)^{2k}}=\sum_{n\in \mathbb{Z} \setminus \{0\}} \frac{1}{n^{2k}} + 2\sum_{n=1}^\infty \sum_{m\in \mathbb{Z}} \frac{1}{(m+n\tau)^{2k}}.$$
The first sum is $2\zeta(2k)$. For the double sum, using a product expansion for $\sin(\pi\tau)$ and a Fourier expansion of $\log(\sin(\pi\tau))$ one can show that 
$$G_{2k}(\tau)=\sum_{n\in \mathbb{Z} \setminus \{0\}} \frac{1}{n^{2k}} + 2\sum_{n=1}^\infty \sum_{m\in \mathbb{Z}} \frac{1}{(m+n\tau)^{2k}}=2\zeta(2k)+2\frac{(2\pi i)^{2k}}{(2k-1)!} \sum_{m\geq 1} \sum_{r|m} r^{2k-1}e^{2\pi i m \tau}.$$
A proof of this alternative way to write $G_{2k}(\tau)$ can be found in Silverman's ``Advanced topics in the arithmetic of elliptic curves'', Chapter I, Lemma 7.1.1. (and specifically the end of Proposition 7.1 in page 57). In your case, 
$$G_{4}(i)=\sum_{n\in \mathbb{Z} \setminus \{0\}} \frac{1}{n^{4}} + 2\sum_{n=1}^\infty \sum_{m\in \mathbb{Z}} \frac{1}{(m+ni)^{4}} = 2\zeta(4) + 2\frac{(2\pi )^{4}}{3!} \sum_{m\geq 1} \sum_{r|m} r^{3}e^{-2\pi m},$$
which is clearly positive. 
