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 $$\int\limits_{0}^\infty g(t)dt=\lim\limits_{k\to\infty}\lim\limits_{m\to\infty}\sum\limits_{n=1}^{2^mk}2^{-m}g(n2^{-m})=\lim\limits_{m\to\infty}\lim\limits_{k\to\infty}\sum\limits_{n=1}^{2^mk}2^{-m}g(n2^{-m})=\lim\limits_{m\to\infty}\sum\limits_{n=1}^{\infty}2^{-m}g(n2^{-m})$$
• It's $2^{-k} g(n 2^{-k})$. You wrote it with all n's. – el_tenedor Sep 18 '16 at 8:32
• Yes! Thanks I was writing the wrong thing. I think it's right now- $2^{-m}$ will be the length of the interval, which we are letting go to 0. – xavier17 Sep 18 '16 at 8:36