# Uniformly continuous and integrable function $f:[0,\infty)\to (0,\infty)$ with $\lim_{t\to\infty}f(t)=C$ satisfies $\sum_{n=1}^\infty f(n)=\infty$

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!

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

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 $$\int_{0}^{\infty} f(x)dx \ge \sum_{n} 0.5 \cdot f(n+0.5) = \sum_{n} 0.5 \cdot \frac{1}{n} = \infty$$
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.