Putnam 2010, Problem A-6

Let \( f : [0, \infty) \to \mathbb{R} \) be a strictly decreasing continuous function such that \( \lim_{x \to \infty} f(x) = 0 \). Prove that
\[
\int_0^\infty \frac{f(x) - f(x+1)}{f(x)} \, dx
\]
diverges.

Solution:
**First solution.** Note that the hypotheses on $f$ imply that $f(x) > 0$ for all $x \in [0, +\infty)$, so the integrand is a continuous function of $f$ and the integral makes sense. Rewrite the integral as
$$\int_0^\infty \left(1 - \frac{f(x+1)}{f(x)}\right) dx,$$
and suppose by way of contradiction that it converges to a finite limit $L$. For $n \geq 0$, define the Lebesgue measurable set
$$I_n = \{x \in [0,1] : 1 - \frac{f(x+n+1)}{f(x+n)} \leq 1/2\}.$$

Then $L \geq \sum_{n=0}^\infty \frac{1}{2}(1 - \mu(I_n))$, so the latter sum converges. In particular, there exists a nonnegative integer $N$ for which $\sum_{n=N}^\infty (1 - \mu(I_n)) < 1$; the intersection
$$I = \bigcup_{n=N}^\infty I_n = [0,1] - \bigcap_{n=N}^\infty ([0,1] - I_n)$$
then has positive Lebesgue measure.

By Taylor's theorem with remainder, for $t \in [0,1/2]$,
$$-\log(1-t) \leq t + \frac{t^2}{2} \sup_{t \in [0,1/2]} \left\{\frac{1}{(1-t)^2}\right\}$$
$$= t + 2t^2 \leq 2t.$$

For each nonnegative integer $n \geq N$, we then have
$$L \geq \int_N^n \left(1 - \frac{f(x+1)}{f(x)}\right) dx$$
$$= \sum_{i=N}^{n-1} \int_0^1 \left(1 - \frac{f(x+i+1)}{f(x+i)}\right) dx$$
$$\geq \sum_{i=N}^{n-1} \int_{I} \left(1 - \frac{f(x+i+1)}{f(x+i)}\right) dx$$
$$\geq \frac{1}{2} \sum_{i=N}^{n-1} \int_{I} \log \frac{f(x+i)}{f(x+i+1)} dx$$
$$= \frac{1}{2} \int_{I} \left(\sum_{i=N}^{n-1} \log \frac{f(x+i)}{f(x+i+1)}\right) dx$$
$$= \frac{1}{2} \int_{I} \log \frac{f(x+N)}{f(x+n)} dx.$$

For each $x \in I$, $\log f(x+N)/f(x+n)$ is a strictly increasing unbounded function of $n$. By the monotone convergence theorem, the integral $\int_I \log(f(x+N)/f(x+n)) dx$ grows without bound as $n \to +\infty$, a contradiction. Thus the original integral diverges, as desired.

**Remark.** This solution is motivated by the commonly-used fact that an infinite product $(1+x_1)(1+x_2)\cdots$ converges absolutely if and only if the sum $x_1 + x_2 + \cdots$ converges absolutely. The additional measure-theoretic argument at the beginning is needed because one cannot bound $-\log(1-t)$ by a fixed multiple of $t$ uniformly for all $t \in [0,1)$.

Greg Martin suggests a variant solution that avoids use of Lebesgue measure. Note first that if $f(y) > 2f(y+1)$, then either $f(y) > \sqrt{2}f(y+1/2)$ or $f(y+1/2) > \sqrt{2}f(y+1)$, and in either case we deduce that
$$\int_{y-1/2}^{y+1/2} \frac{f(x) - f(x+1)}{f(x)} dx > \frac{1}{2}\left(1 - \frac{1}{\sqrt{2}}\right) > \frac{1}{7}.$$

If there exist arbitrarily large values of $y$ for which $f(y) > 2f(y+1)$, we deduce that the original integral is greater than any multiple of $1/7$, and so diverges. Otherwise, for $x$ large we may argue that
$$\frac{f(x) - f(x+1)}{f(x)} > \frac{3}{5} \log \frac{f(x)}{f(x+1)}$$
as in the above solution, and again get divergence using a telescoping sum.

**Second solution.** (Communicated by Paul Allen.) Let $b > a$ be nonnegative integers. Then
$$\int_a^b \frac{f(x) - f(x+1)}{f(x)} dx = \sum_{k=a}^{b-1} \int_0^1 \frac{f(x+k) - f(x+k+1)}{f(x+k)} dx$$
$$= \int_0^1 \sum_{k=a}^{b-1} \frac{f(x+k) - f(x+k+1)}{f(x+k)} dx$$
$$\geq \int_0^1 \sum_{k=a}^{b-1} \frac{f(x+k) - f(x+k+1)}{f(x+a)} dx$$
$$= \int_0^1 \frac{f(x+a) - f(x+b)}{f(x+a)} dx.$$

Now since $f(x) \to 0$, given $a$, we can choose an integer $l(a) > a$ for which $f(l(a)) < f(a+1)/2$; then $\frac{f(x+a) - f(x+l(a))}{f(x+a)} \geq 1 - \frac{f(l(a))}{f(a+1)} > 1/2$ for all $x \in [0,1]$. Thus if we define a sequence of integers $a_n$ by $a_0 = 0$, $a_{n+1} = l(a_n)$, then
$$\int_0^\infty \frac{f(x) - f(x+1)}{f(x)} dx = \sum_{n=0}^\infty \int_{a_n}^{a_{n+1}} \frac{f(x) - f(x+1)}{f(x)} dx$$
$$> \sum_{n=0}^\infty \int_0^1 (1/2) dx,$$
and the final sum clearly diverges.

**Third solution.** (By Joshua Rosenberg, communicated by Catalin Zara.) If the original integral converges, then on one hand the integrand $(f(x) - f(x+1))/f(x) = 1 - f(x+1)/f(x)$ cannot tend to 1 as $x \to \infty$. On the other hand, for any $a \geq 0$,
$$0 < \frac{f(a+1)}{f(a)}$$
$$< \frac{1}{f(a)} \int_a^{a+1} f(x) dx$$
$$= \frac{1}{f(a)} \int_a^\infty (f(x) - f(x+1)) dx$$
$$\leq \int_a^\infty \frac{f(x) - f(x+1)}{f(x)} dx,$$
and the last expression tends to 0 as $a \to \infty$. Hence by the squeeze theorem, $f(a+1)/f(a) \to 0$ as $a \to \infty$, a contradiction.