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Let $f:[0,\infty)\to (0,\infty)$ be a uniformly continuous function with $\lim_{t\to\infty}f(t)=C.$ Prove that $\int_0^\infty f(t)dt=\infty$ $\Longleftrightarrow$ $\sum_{n=1}^\infty f(n)=\infty$.

My attempt

For the case $C>0$, I can prove that both $\int_0^\infty f(t)dt=\infty$ and $\sum_{n=1}^\infty f(n)=\infty$ hold. The difficulty is in the case that $C=0$ and I got stuck. Any helps would be highly appreciated!

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    $\begingroup$ Of course $C=0$ is the only case where "uniformly continuous" will be needed. $\endgroup$
    – GEdgar
    Oct 28, 2021 at 16:07

2 Answers 2

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I think the statement does not hold.

Take $f(x)$ to be a piecewise linear "bump" function with $f(n) = \frac{1}{n^2}$ but $f(n+0.5) = \frac{1}{n}$ for all $n$.

Then $f$ vanishes at $\infty$ and is uniformly continuous as a continuous function that vanishes at $\infty$.

We get $$\sum_{n} f(n) = \sum_{n} \frac{1}{n^2}< \infty$$ But $$\int_{0}^{\infty} f(x)dx \ge \sum_{n} 0.5 \cdot f(n+0.5) = \sum_{n} 0.5 \cdot \frac{1}{n} = \infty$$

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As suggested by @Claudio Moneo, this isn't true all time. But making some changes in conditions of statement we can derive some analogous case to this. Let $f$ be a continuous function such that, $$f:[0,\infty)\rightarrow (0,\infty)$$ $$\lim_{n\to\infty}nf(n)= A$$ $f'(x)$ exists And if $$\int_{0}^{\infty}f(t)dt$$ diverges Then all these conditions imply that, $$\sum_{n\in\mathbb{N}}f(n)\rightarrow \infty$$ $\textbf{PROOF}:-$ This is a direct result obtained via partial summation. Consider the integral, $$\int_{0}^{\infty}f(x)dx$$ Which diverges by our conditions. Integration by parts yields, $$[f(x)]_{x\to\infty}-[f(x)]_{x\to 0}-\int_{0}^{\infty}xf'(x)dx$$ As per the conditions above the limit $\lim_{x\to\infty}xf(x)$ converges. And $\lim_{x\to 0}xf(x)$ is $0$ because $f(0)$ converges. So coming to integral, Abel summation converts the integral to summation, $$\int_{0}^{\infty}xf'(x)dx=\sum_{n\ge 1}A(n-1)f(n)-\sum_{n\ge 1}A(n)f(n)$$, Where, $A(u)=\sum_{n\le u, n\in\mathbb{N}}1$. Combining the both sums because they are between same bounds leads to, $$-\sum_{n\ge 1}f(n)$$. Placing this in place of that integrand leads to, $$A-\int_{0}^{\infty}xf'(x)dx=A+\sum_{n\ge 1}f(n)$$ Now since the integral diverges, $\sum_{n\ge 1}f(n)$ also diverges. If we proceed in the reverse, we may also prove that the converse also holds.

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