Calculate $\lim{_{n\rightarrow \infty}} \frac{1}{n} \int_0^n f(x) dx$

Question

If $f: \mathbb{R} \to (0, \infty)$ is continuous and $$\lim_{x\to \infty}\frac{f(x)}{f(x+1)}=0$$ then calculate $$\lim_{n\to \infty} \frac{1}{n} \int_0^n f(x) dx$$

I tried solving by writing the definition of continuity but didn't find the right way to finish it, also tried taking logarithm, no result there neither.

Answer 1

The condition that $f(x)/f(x+1) \to 0$ means that, eventually, $f(x+1)$ is significantly larger than $f(x)$ which is effectively a rapid growth condition that is the opposite of what is required for the integral to converge.

The stated condition implies there is a whole number $M$ with the property that where $x > M$ implies $\dfrac{f(x)}{f(x+1)} \le \frac 12$, or equivalently $f(x+1) \ge 2 f(x)$.

Let $a_n = \min\{ f(x) : M+n \le x \le M+n+1\}$. Since $f$ is positive valued and continuous you have $a_0 > 0$, and if $x \in [M+n,M+n+1]$ then $$f(x) \ge 2 f(x-1) \ge a_{n-1}$$ and by taking the minimum over all $x$ in the interval you find $a_n \ge 2 a_{n-1}$.

You then obtain $$\int_M^{M+n} f(x) \, dx = \sum_{k=0}^{n-1} \int_{M+k}^{M+k+1} f(x) \, dx \ge \sum_{k=0}^{n-1} a_k = (2^n-1)a_0$$

for all $n$. From here you should be able to verify $$\frac 1n \int_0^n f(x) \, dx = \infty.$$

Answer 2

The limit actually implies that the function diverges to $\infty$. This is itself enough to say that the limit diverges, but it's actually the speed at which it grows that causes the limit to diverge. The limiting property of $f$ tells us that for large enough $x$, we may take

$$\frac{f(x)}{f(x+1)}<\frac{1}{2}\implies f(x+1)>2f(x)$$

Generalizing this yields

$$f(x+k)>2^kf(x)$$

for large values of $x$. Let's assume that for some critical $N\in\mathbb{N}$ and $x\geq N$, $f$ satisfies this property. For any $n>N$ we have

$$\int_0^nf(x)dx=\int_0^Nf(x)dx+\sum_{k=1}^{n-N}\int_{N+k-1}^{N+k}f(x)dx=\int_0^Nf(x)dx+\sum_{k=1}^{n-N}\int_0^1f(x+N+k-1)dx$$

$$\geq\int_0^Nf(x)dx+\sum_{k=1}^{n-N}2^{k-1}\int_0^1f(x+N)dx=\int_0^Nf(x)dx+\frac{2^{n-N+1}-1}{2-1}\int_0^1f(x+N)dx$$

where we have used the property of geometric series to evaluate the sum. From here, we divide by $n$ and take the limit. The first integral is constant, so it limits to $0$. This leaves us with

$$\lim_{n\to\infty}\frac{1}{n}\int_0^nf(x)dx=\left(\int_0^1f(x+N)dx\right)\lim_{n\to\infty}\frac{2^{n+1-N}-1}{n}$$

This limit diverges to infinity, and we are done.