Find an example such that the sequence $(f(n)-f(n+1))$ is divergent.

Question

Let $f:(0,\infty)\rightarrow\mathbb{R}$ be a function.

My question is Find an example such that the sequence $(f(n)-f(n+1))$ is divergent as $n\to\infty$.

I found a lot of well known functions which they are divergent, but $(f(n)-f(n-1))$ is convergent such as $f(x)=\sqrt{x}$. Even I dealt to some functions with mentioned property but they are cooked up from other simple functions. anyone can take a well known example about my question. Thanks.

Answer 1

What about the function $f(x)=\cos(\pi x)$?

Answer 2

I think $f=e^x$ does the job. The intuition is that the exponential function grows really fast, so that you have: $f(n+1)-f(n) = e^n(e-1) \to \infty$. The problem with $f(x)=\sqrt{x}$ is that it grows slower and slower for $x \to \infty$. Actually even $f(x) = x^2$ satisfies $f(n+1)-f(n) \to \infty$.

I think it can be proven that if $\frac{\mathrm{d}f}{\mathrm{d}x} \to \infty$ when $x \to \infty$, then $f(n+1)-f(n) \to \infty$ when $n \to \infty$.

Answer 3

Suppose $f(n) = \sum_{k=1}^{n}k $, then $f(n) - f(n-1)= n$ (simply you can think $f(n) \propto n^2$ and their difference $f(n) - f(n-1) = f'$ )

I think a similar way, you can find many many examples!

Answer 4

One thing to keep in mind in a question like this is that you can choose the values of $f$ at different numbers however you want! You aren't required to have any nice general formula for the function; you can just define each value separately.

So, for instance, you might start by declaring that $f(1)=0$, say. Then you can pick $f(2)$ to be any real number, to make $f(1)-f(2)$ whatever you want. Then you can pick $f(3)$ to be any real number, and so you can again make $f(2)-f(3)$ whatever you want. Then you can pick $f(4)$ to make $f(3)-f(4)$ be whatever you want. Continuing to pick values $f(n)$ one by one, you can see that you can choose $f(n)-f(n+1)$ to be any sequence whatsoever! To get a function $f:(0,\infty)\to\mathbb{R}$ as the question requires, you then just need to also define $f(x)$ when $x\in(0,\infty)$ is not an integer, which you can do however you like.