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

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.

• This limit should diverge to infinity. It's pretty easy to show via $\epsilon-\delta$ – Don Thousand Nov 5 '18 at 20:37
• what role does $x$ plays in the integral? since you consider a definite integral the result will be $F(n)-F(0)$ . Shouldn't $n$ tend to infinity? – Chaos Nov 5 '18 at 20:51

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 $$\frac 1n \int_0^n f(x) \, dx = \infty.$$

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.