The left cosets $\Gamma_\infty\backslash\Gamma(1)$ correspond to primitive bottom rows $(c,d)\in\mathbb Z^2$, up to simultaneous sign. Put
$$
r=2\operatorname{Re}s+k.
$$
For $\tau=x+iy$, the identities
$$
\operatorname{Im}(\gamma\tau)=\frac{y}{|c\tau+d|^2},
\qquad
j(\gamma,\tau)=c\tau+d
$$
show that the absolute value of a summand is
$$
y^{\operatorname{Re}s}|c\tau+d|^{-r}.
$$
The hypothesis $\operatorname{Re}s>(2-k)/2$ says precisely that $r>2$.

If $\tau$ ranges over a compact subset $K\subset\mathfrak h$, the positive-definite quadratic form $|c\tau+d|^2$ has a uniform lower bound
$$
|c\tau+d|^2\geq C_K(c^2+d^2)
$$
for some $C_K>0$. The summands are therefore bounded uniformly on $K$ by a constant times
$$
(c^2+d^2)^{-r/2}.
$$
The corresponding two-dimensional <lattice sum> converges for $r>2$. The <Weierstrass M-test> proves absolute and locally uniform convergence. This is the <absolute convergence of a weight-k real-analytic Eisenstein series>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 137 / 4 / a / Solution

The absolute value of the summand indexed by a primitive bottom row $(c,d)$ is
$$
y^{\operatorname{Re}s}|c\tau+d|^{-2\operatorname{Re}s-k}.
$$
The exponent exceeds two, so comparison with the lattice sum over $(c,d)\in\mathbb Z^2\setminus\{0\}$ proves absolute and locally uniform convergence.


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