Okay.
So if $S_n = 1-\sum_{m=1}^n \frac 1{4^n}$ its easy to see that $S_{k+1} = S_k -\frac 1{4^{k-1}} < S_k$ so $S_1 >S_2 > S_3.....$.
So every $S_k \le S_1$ and so $\sup S_n = \max S_n = S_1$.
And because, for every $S_k$ there is always $S_{k+1} < S_k$ we can't have have a smallest $S_k$ so $\min S_n$ can not exists.
So what remains to answer are: Are the $S_k$ bounded below and if so what is $\inf S_n$?
Are you familiar with geometric series?
$1 + r + r^2 + r^3 +........ + r^n = M_n$
Then $M_n(1-r) = (1 + r + r^2 + r^3 +........ + r^n) - (r+r^2 + ...... + r^n + r^{n+1}) = 1-r^{n+1}$ so
$M_n = \frac {1-r^{n+1}}{1-r}$.
So $\sum_{m=1}^n \frac 14 = (\sum_{m=0}^n \frac 14) -1=\frac {1-(\frac 14)^{n+1}}{1-\frac 14} -1=$
$\frac {4(1-(\frac 14)^{n+1}}3 -1= \frac {4-(\frac 14)^n}3 - 1=\frac {1-\frac 1{4^n}}3$
So $S_n = 1-\frac {1-\frac 1{4^n}}3= \frac {2+\frac 1{4^n}}3$.
And as $\frac 1{4^n} >0$ we can see than indeed $S_n > \frac 23$ and the sequence is bounded below by $\frac 23$.
If we take any number larger than $\frac 23$, say $\frac 23 + \iota$ then we can find a $\frac 1{4^n}$ so that $\frac 23 < \frac {2+\frac 1{4^n}}3 < \frac 23 + \iota$. (Why?)
So any number larger than $\frac 23$ is not a lower bound, so $\frac 23$ is the greatest lower bound.
And $\inf S_n = \frac 23$.