I have the following true/false question

If $f:[a,\infty)\to\mathbb{R}$ is monotonically decreasing, and the integral $\int_{1}^{\infty}f(x)dx$ is convergent, then $\underset {x\to\infty} \lim f(x)=0$.

My intuition is that it's true but not sure how to prove it or if actually there is a counter example.


  • $\begingroup$ One possible starting point: Case 1: $f$ is unbounded. Case 2: $f$ is bounded. $\endgroup$ – Arthur Aug 2 '17 at 9:30
  • 2
    $\begingroup$ Assume $f$ is bounded. It is decreasing, and so the limit $\lim_{x\to \infty} f(x)$ exists. Say $\lim_{x\to \infty} f(x) = a >0$. But then $\int_1^\infty f(x) \,dx \geq a \int_1^\infty \, dx = \infty$. $\endgroup$ – Teddy Aug 2 '17 at 9:39

It is true. Assume $f:[1,\infty)\to \mathbb R$ is monotonically decreasing and prove $$ \int_1^\infty f(x)~dx \text{ convergent }\Rightarrow \lim_{x\to \infty} f(x)= 0 $$ by contraposition.

Let be $\lim f(x)\neq 0$. If $f(x)\to -\infty$ then $\int_1^\infty f(x)~dx=-\infty$ is obvious. Otherwise let be $c:=\lim_{x\to\infty} f(x)\in\mathbb R\setminus\{0\}$.

First, $c>0$. Then we use $f(x)\geq c$ and conclude $$ \int_1^\infty f(x)~dx=\lim_{R\to\infty}\int_1^Rf(x)~dx\geq \lim_{R\to\infty}\int_1^Rc~dx=\lim_{R\to\infty}(R-1)c=\infty. $$

Next, $c<0$. Since $f(x)\to c$ we get $X>0$ such that $f(x)\leq \frac12c$ for all $x\geq X$ and conclude \begin{align} \int_1^\infty f(x)~dx&=\int_1^Xf(x)~dx+\int_X^\infty f(x)~dx=\int_1^X f(x)~dx+\lim_{R\to\infty}\int_X^Rf(x)~dx\\ &\geq \int_1^Xf(x)~dx+\lim_{R\to\infty}\int_1^R\frac12c~dx\\ &=\int_1^Xf(x)~dx+\lim_{R\to\infty}\frac12(R-X)c=\infty. \end{align}


From the convergence of the integral, for each $\epsilon>0$ there exists $x_0>a$ such that $\left|\int_u^vf(x)\,\mathrm dx\right|<\epsilon$ for all $v>u>x_0$. Using monotonicity, $(v-u)f(u)\ge \int_u^vf(x)\,\mathrm dx\ge (v-u)f(v)$. Letting $u=v-1$, conclude that $f(v)\le \epsilon$ for $v>x_0+1$. Also, transform to $f(u)>\frac\epsilon{v-u}$ and conclude $f(u)\ge 0$ by letting $v\to+\infty$.


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