Why is the modular $\lambda$ function a quotient of two meromorphic functions in the U.H.P.? In Ahlfors' complex analysis text, page 278 it says:

It is quite clear from (9) that $\lambda(\tau)$ is the quotient of two analytic functions in the upper half plane $\text{Im} \tau > 0$.

Formula (9) is the definition of the Weierstrass $\wp$ function $$\wp(z;\omega_1, \omega_2)=\frac{1}{z^2}+ \sum_{\omega \neq 0} \frac{1}{(z-\omega)^2}-\frac{1}{\omega^2} $$
where the sum is taken over all linear combinations $$\omega=n_1 \omega_1+n_2 \omega_2 $$ with integer coefficients (except the zero combination).
The $\lambda$ function is defined as $$\lambda(\tau)= \frac{e_3-e_2}{e_1-e_2} $$ where $e_1=\wp(\omega_1/2),e_2=\wp(\omega_2/2),e_3=\wp((\omega_1+\omega_2)/2)$, and $\tau$ is the ratio $\omega_2/\omega_1$.
My attempt:
Following the author's advice I found that $\lambda$ equals
$$\frac{\frac{1}{\left( \frac{\omega_1+\omega_2}{2} \right)^2}+\sum \left[ \frac{1}{\left( \frac{\omega_1+\omega_2}{2}-n_1 \omega_1-n_2 \omega_2 \right)^2}-\frac{1}{\left( n_1 \omega_1+n_2 \omega_2 \right)^2} \right]-\frac{1}{\left( \frac{\omega_2}{2} \right)^2}- \sum \left[ \frac{1}{\left( \frac{\omega_2}{2}-n_1 \omega_1-n_2 \omega_2 \right)^2}-\frac{1}{\left( n_1 \omega_1+n_2 \omega_2 \right)^2}\right]}{\frac{1}{\left( \frac{\omega_1}{2} \right)^2}+\sum \left[ \frac{1}{\left( \frac{\omega_1}{2}-n_1 \omega_1-n_2 \omega_2 \right)^2}-\frac{1}{\left( n_1 \omega_1+n_2 \omega_2 \right)^2} \right]-\frac{1}{\left( \frac{\omega_2}{2} \right)^2}-\sum \left[ \frac{1}{\left( \frac{\omega_2}{2}-n_1 \omega_1-n_2 \omega_2 \right)^2}-\frac{1}{\left( n_1 \omega_1+n_2 \omega_2 \right)^2} \right]}  $$
multiplying both the numerator and the denominator by $\omega_1^2$ we get the numerator $$ \frac{1}{\left( \frac{1+\tau}{2} \right)^2}+\sum \left[ \frac{1}{\left( \frac{1+\tau}{2}-n_1-n_2 \tau \right)^2}-\frac{1}{\left( n_1 +n_2 \tau \right)^2} \right]-\frac{1}{\left( \frac{\tau}{2} \right)^2}- \sum \left[ \frac{1}{\left( \frac{\tau}{2}-n_1-n_2 \tau \right)^2}-\frac{1}{\left( n_1 +n_2 \tau \right)^2}\right]$$ and the denominator
$$\frac{1}{\left( \frac{1}{2} \right)^2}+\sum \left[ \frac{1}{\left( \frac{1}{2}-n_1 -n_2 \tau \right)^2}-\frac{1}{\left( n_1+n_2 \tau \right)^2} \right]-\frac{1}{\left( \frac{\tau}{2} \right)^2}-\sum \left[ \frac{1}{\left( \frac{\tau}{2}-n_1-n_2 \tau \right)^2}-\frac{1}{\left( n_1 +n_2 \tau \right)^2} \right] $$
I think the next step is proving that both the numerator and denominator are analytic as a function of $\tau$ in $\Im \tau>0$. Firstly, I don't know how to do that, and secondly, shouldn't they be analytic for $\Im \tau<0$ as well?
Could anyone please tell me why are the numerator and denominator analytic in $\Im \tau>0$?
 A: 
I think the next step is proving that both the numerator and denominator are analytic as a function of $\tau$ in $\Im \tau > 0$. Firstly, I don't know how to do that, and secondly, shouldn't they be analytic for $\Im \tau < 0$ as well?

Yes, the natural way is to see that numerator and denominator are analytic functions of $\tau$.
Let me answer the second point first, since that is short and easy: Yes, they are analytic for $\Im \tau < 0$ too, but we're not interested in that. It is of course an arbitrary choice whether to use the upper or the lower half plane, but the restriction to a half plane is necessary, since the real line is a natural boundary of $\lambda$. If you look at the sums
$$\sum' \frac{1}{(\rho - n_1 - n_2\tau)^2} - \frac{1}{(n_1+n_2\tau)^2},$$
where $\rho\in \left\lbrace\dfrac12,\,\dfrac{\tau}{2},\,\dfrac{1+\tau}{2} \right\rbrace$, you see that each term is analytic in the entire upper (and lower) half plane, but infinitely many terms have a pole in each $\tau \in \mathbb{Q}$, and that means the sum cannot be well-behaved in any interval of $\mathbb{R}$.
So let's come to the first point, how to see that these are analytic functions of $\tau$ (in the upper half plane). Every term of the sums is analytic, so the standard way of seeing analyticity is to show that the sums converge locally uniformly in the half plane. That is done quite similarly to showing the locally uniform convergence of $\wp$.
For $\omega = n_1 + n_2\tau \in \Omega\setminus \{0\}$, we have
$$\frac{1}{(\rho-\omega)^2} - \frac{1}{\omega^2} = \frac{\rho(2\omega-\rho)}{(\rho-\omega)^2\omega^2}.$$
For $\lvert\tau\rvert \leqslant K$ and $\lvert \omega\rvert \geqslant 2K$ (with $K\geqslant 1$), we can estimate the absolute modulus by
$$\frac{M}{\lvert\omega\rvert^3},$$
since $\lvert \rho-\omega\rvert \geqslant \frac12\lvert\omega\rvert$ and $\lvert 2\omega-\rho\rvert \leqslant 3\lvert\omega\rvert$. Now bounding $\lvert n_1 + n_2\tau\rvert^2$ below by a positive multiple of $n_1^2 + n_2^2$ will establish locally uniform convergence by Weierstraß' $M$-test. To get such an estimate, one must keep $\tau$ away from the real line, $\Im\tau \geqslant \varepsilon > 0$. Writing $\tau = \xi + i\eta$, we have
$$\lvert n_1 + n_2\tau\rvert^2 = (n_1+n_2\xi)^2 + n_2^2\eta^2 \geqslant (n_1+n_2\xi)^2 + \varepsilon^2 n_2^2,$$
and the compactness of the unit circle ensures the existence of a $c > 0$ with
$$(x+y\xi)^2 + \varepsilon^2y^2 \geqslant c\cdot (x^2+y^2)$$
for all real $x,y$. Thus the sums converge uniformly on every compact subset of the upper half plane, hence yield analytic functions.
