Show that $G_4(i)\neq 0$, and $G_6(\rho)\neq 0$, $\rho=e^{2\pi i /3}$ Let $G_k$ denote the Eisenstein series of weight $k$. I know that $G_k(i)=0$ if $k \not\equiv 0 \ (mod \  4)$ and $G_k(\rho)=0$ if $k \not\equiv 0 \ (mod \  6)$. However, I want to know how to show, that $G_4(i)\neq0$ and $G_6(\rho)\neq0$, without using the $\frac{k}{12}$-formula.
 A: Here's a direct proof that $G_4(i) \neq 0$ that uses only the series.  
The point is simply to estimate the tail $|a|,|b| \geq N$  and then calculate the first few terms for $|a|,|b| < N$.  In fact, using very simply bounds we can take $N=2$.
First note that $|a|+|b| \leq \sqrt{3}|a+bi|$ which follows from
$$ 2|a||b| \leq 2 (a^2+b^2) \implies a^2+b^2 + 2|a||b| \leq 3(a^2+b^2) \implies (|a|+|b|)^2 \leq 3(a^2+b^2). $$
This with integral comparison, easily gives  \begin{align*} \left|\sum_{|a|,|b| \geq N} \frac{1}{(a+bi)^4} \right| &\leq \sum_{|a|,|b| \geq N} \frac{1}{|a+bi|^4} \\\\ &\leq \sqrt{3}^4 \sum_{|a|,|b| \geq N} \frac{1}{(|a|+|b|)^4} \\\\ &= 9 \cdot 4 \sum_{a,b \geq N} \frac{1}{(a+b)^4} \\\\ &= 36 \sum_{a \geq N} \sum_{b \geq N} \frac{1}{(a+b)^4} \\\\ &= 36 \sum_{a \geq N} \sum_{b \geq N+a} \frac{1}{b^4} \\\\ &\leq 36 \sum_{a \geq N} \int_{N+a-1}^{\infty} \frac{1}{x^4}dx \\\\&= 12 \sum_{a \geq N} \frac{1}{(N+a-1)^3} \\\\&= 12 \sum_{a \geq 2N-1} \frac{1}{a^3} \\\\ &\leq 12 \int_{2N-2}^{\infty} \frac{1}{x^3} dx  \\\\ &= \frac{12}{(2N-2)^2} \\\\ &= \frac{3}{(N-1)^2}.  \end{align*} 
Thus, when $N=2$, the tail is bounded by $\Large{\color{red}{3}}$.  
Now, since conjugation permutes $\mathbb{Z} + i\mathbb{Z}$, we know that $G_4(i)$ is real.  Thus, we need only worry about the real part of the sum.  We have $$\text{Re} \left(\sum_{\substack{|a|,|b| < N \\ (a,b) \neq (0,0) }} \frac{1}{(a+bi)^4} \right) = \sum_{\substack{|a|,|b| < N \\ (a,b) \neq (0,0) }} \frac{a^4+b^4-6a^2b^2}{(a^2+b^2)^4}.$$  With $N=2$, a very short calculation gives $\Large{\color{red}{5}}$, a positive number.  Note that there is no way that the tail will make it negative.  Done.
