Did I get the sum of this geometric sequence right? (Last few terms fix is geometric now)
$$ \frac{1}{(2^{k})^2}+\frac{1}{(2^{(k-1)})^2}+\dots+\frac{1}{(2^2)^2}+\frac{1}{(2)^2}+1 $$
Using
$S_n = \frac{a_1(r^n-1)}{r-1}$ with $a_1 =1,r=\frac{1}{2^2}, n=k \Rightarrow S_n = \frac{4-4^{1-k}}{3}$
A sum such as $$\frac1{2^{k^2}} +\frac1{2^{(k-1)^2}} +\frac1{2^{(k-2)^2}} +\frac1{2^{(k-3)^2}} +\ldots +\frac1{2^2} +1$$ for positive $k$ isn't a geometric series because the exponents don't follow an arithmetic progression. Also, how come the second-to-last term is $1/2^2?$