Sequence of series converges to improper integral Let $g \colon [0,\infty) \to \mathbb{R}$ be a monotonous function.
Suppose $g$ only attains positive values and is (not necessarily strictly) decreasing.
Does the sequence of series
$$
s_k := \sum_{n=1}^\infty 2^{-k} g(n 2^{-k})
$$
converge to
$$
\int_0^\infty g(t) dt
$$
for $k \to \infty$?
Since $g$ is positive, the series either converges absolutely or "converges" to $\infty$.
Since we always have $s_k \leq \int_0^\infty g(t) dt$ let's assume that the series converges for all $k \in \mathbb{N}$.
Note that the series approximates the improper integral like Riemannian sums with interval length $2^{-k}$.

I believe the answer is yes, if, as you said, the integral exists.  By definition, $\int\limits_0^\infty g(t)dt=\lim\limits_{k\to\infty}\int\limits_{0}^kg(t)dt$.  If we assume this integral exists, then we can take the limit over $k\in\mathbb{Z}$, and hence choose $k$ a positive integer.  Then we have
$$
\int\limits_{0}^{k}g(t)dt=\lim\limits_{m\to\infty}\sum\limits_{n=1}^{2^mk}2^{-m}g(n2^{-m})
$$
It follows that
$$
\int\limits_{0}^\infty g(t)dt=\lim\limits_{k\to\infty}\lim\limits_{m\to\infty}\sum\limits_{n=1}^{2^mk}2^{-m}g(n2^{-m})=\lim\limits_{m\to\infty}\lim\limits_{k\to\infty}\sum\limits_{n=1}^{2^mk}2^{-m}g(n2^{-m})=\lim\limits_{m\to\infty}\sum\limits_{n=1}^{\infty}2^{-m}g(n2^{-m})
$$
