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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 $2^{-k}$.

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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 $$ \int\limits_{0}^\infty g(t)dt=\lim\limits_{k\to\infty}\lim\limits_{m\to\infty}\sum\limits_{n=1}^{2^mk}2^{-m}g(n2^{-m})=\lim\limits_{m\to\infty}\lim\limits_{k\to\infty}\sum\limits_{n=1}^{2^mk}2^{-m}g(n2^{-m})=\lim\limits_{m\to\infty}\sum\limits_{n=1}^{\infty}2^{-m}g(n2^{-m}) $$

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  • $\begingroup$ Are you sure you did't mix up m's and k's in the series? $\endgroup$ – el_tenedor Sep 18 '16 at 8:25
  • $\begingroup$ I meant n's and k's. $\endgroup$ – el_tenedor Sep 18 '16 at 8:26
  • $\begingroup$ Hmm.... where? I can't spot anything. I certainly might have. $\endgroup$ – xavier17 Sep 18 '16 at 8:28
  • $\begingroup$ It's $2^{-k} g(n 2^{-k})$. You wrote it with all n's. $\endgroup$ – el_tenedor Sep 18 '16 at 8:32
  • $\begingroup$ 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. $\endgroup$ – xavier17 Sep 18 '16 at 8:36

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