Let
$$f(t):=e^{-t^2/4}, \ \ \ t\ge0$$ (which is decreasing and uniformly continuous)
and $$f_n(t) := \sum_{k=0}^{n 2^n} \exp \left[ - \frac{1}{4} \exp \left( \frac{(k+1)}{2^n} \right)^2 \right] 1_{[k2^{-n},(k+1)2^{-n})}(t).$$
How can I show that
$$f_n(t)\le f(t) \ \ \ \forall t\ge 0$$
and
$$\lim_{n \to \infty}f_n(t)=f(t)$$
I'm assuming there was a misprint and it should be
$$f_n(t) = \sum_{k=0}^{n 2^n} \exp \left[ - \frac{1}{4} \left( \frac{(k+1)}{2^n} \right)^2 \right] 1_{[k2^{-n},(k+1)2^{-n})}(t).$$
Note that this is just another way to write the step function
$$f_n(t) = \begin{cases} \exp \left[ - \frac{1}{4} \left( \frac{(k+1)}{2^n} \right)^2 \right] & \text{if $t\in I_{n,k}$ for some $0\le k\le n 2^n$}, \\ 0 & \text{otherwise,} \end{cases}$$
where the intervals are given by $I_{n,k} = \left[\frac{k}{2^n},\frac{(k+1)}{2^n}\right)$.
Using $f(t) = e^{-t^2/4}$ we can also write this as $$ f_n(t) = \begin{cases} f\left(\frac{(k+1)}{2^n} \right) & \text{if $t\in I_{n,k}$ for some $0\le k\le n 2^n$}, \\ 0 & \text{otherwise,} \end{cases} $$
Hence $f_n(t)$ is obtained on $\left[0, n+\frac{1}{2^n}\right)=\bigcup_{k=0}^{n2^n} I_{n,k}$ by evaluating $f$ at the right bound of the interval containing $t$.
Since $f$ is decreasing, we immediately get that $f_n(t) \le f(t)$ whenever $t$ is in one of the intervals: for $t\in I_{n,k}$ we have $t < \frac{k+1}{2^n}$, hence $$ f_n(t) = f\left( \frac{k+1}{2^n}\right) \le f(t). $$ Beyond the intervals, that is for $t\ge n+\frac{1}{2^n}$, we have $$ f_n(t) = 0 \le f(t). $$
If we look at the left bound of the intervals instead, we obtain an upper bound. That is, let $t\in I_{n,k}$, then $t\ge \frac{k}{2^n}$ so that $k\le t2^n$ and we get $$ f_n(t) = f\left(\frac{k+1}{2^n}\right) \ge f\left(\frac{t2^n+1}{2^n}\right) = f\left(t+\frac{1}{2^n}\right). $$
In summary, for $0\le t < n+\frac{1}{2^n}$ we have $$ f\left(t+\frac{1}{2^n}\right) \le f_n(t) \le f(t). $$
Taking the limit $n\to\infty$ and using continuity of $f$, the squeeze theorem implies that $$ \lim_{n\to\infty} f_n(t) = f(t) $$ for all $t\ge 0$.