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If $f : [a, \infty) \rightarrow \Bbb R$ is monotonically decreasing and the integral $\int_0^\infty f(x) \,dx$ is convergent, then $\lim_{x\rightarrow\infty} f(x) = 0$. I don't really understand why this is true.

I know that if $f$ is monotonically decreasing then $f$ is bounded ($\exists M\in \Bbb R s.t. f(x)\leq M , \forall n\geq k)$.

I also know that $$\int_0^\infty f(x) \,dx = \lim_{b\rightarrow\infty}\int_0^b f(x) \,dx < \infty$$ I guess I can say something about the the monotonicity of limits, but I don't know how to continue from here. I would love for some help (it doesn't need to be formal proof, I just want to understand it).

Thank you.

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    $\begingroup$ $\sum_{n=0}^\infty f(n) \geq \int_{0}^\infty f(x)dx \geq \sum_{n=1}^\infty f(n)$ $\endgroup$ – Michael Jul 26 '18 at 21:32
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Fix $x_0 > 0$. Since $f$ is decreasing, for any $x \in [nx_0, (n+1)x_0]$ we have $f((n+1)x_0) \le f(x)$.

Therefore

\begin{align} x_0 \sum_{n=1}^\infty f(nx_0) &= x_0 \sum_{n=0}^\infty f((n+1)x_0)\\ &= \sum_{n=0}^\infty \int_{nx_0}^{(n+1)x_0}f((n+1)x_0)\,dx \\ &\le \sum_{n=0}^\infty \int_{nx_0}^{(n+1)x_0}f(x)\,dx\\ &= \int_0^\infty f(x)\,dx \end{align}

so the series $\sum_{n=1}^\infty f(nx_0)$ converges and hence $\lim_{n\to\infty} f(nx_0) = 0$.

Now conclude that $\lim_{x\to\infty} f(x) = 0$.

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It might be best to approach this by contradiction. In particular, if we assume $\lim_{x\to\infty}f(x)\neq 0$, then either the limit is nonzero or $f$ decreases without bound. It should be possible to show in each case that the integral does not converge.

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