# Good examples of violations to the “law of sufficiently large intervals”

Tis a common mistake among us mere mortals (laypeople, students, physicists, and the like) to assume that because a function (or sequence) has a certain property over some sufficiently large interval, it must maintain this property as its argument tends to infinity. Some particularly prominent examples are the "apparent" asymptote, the "looks-like" limit at infinity, and the "almost" Cauchy sequence.

Well, I was trying to explain why this is not the case when I realised that for every example I could think of, the idea of looking at a "sufficiently large" interval did suffice to predict the limiting behaviour of the function. In fact, I've started noticing that even mathematicians will casually use this kind of reasoning when the situation does not require otherwise.

The problem is that most "ordinary" functions have relatively consistent behaviour over their entire domain. If $$f(x)$$ is continuous, increasing, $$f(x)<3$$ for all $$x<10,000$$, and $$f(10,000)=2.999999978$$, then odds are that $$\lim_{x\to\infty}f(x)=3$$. Of course, there might be some absurdly large value where $$f$$ starts increasing much faster but unless it's being done intentionally there isn't.

This brings me to my question.

What are some "natural" examples of functions (or classes thereof) that appear to obey the "law of sufficiently large intervals" but do not?

By "natural" I mean something that might appear in a high-school or undergrad textbook and could plausibly fool a TA if they weren't careful; a definable function whose formula does not immediately spoil the reveal. Obviously, something like...

$$f(x)=\frac{x}{10^{21}}-\frac 1x$$

...won't work because anyone who looks at it more than once will realise that $$f(x)$$ is only "small" when $$x<10^{21}$$.

I'm looking more for examples where, without further work, it still seems reasonable that the limiting behaviour of the function is predicted by its behaviour over a large interval.

• What about the divergence of $\sum \frac 1p$ where the sum is taken over $p$ prime? If you sum the first $10^4$ terms you only get $2.7$, and if you go to $2\times 10^4$ you only get up to $2.77$, and yet the sum diverges....the $k^{th}$ partial sum being on the order of $\log \log k$. – lulu Nov 16 '19 at 0:46
• @lulu I like this example - if you made the claim $\displaystyle\sum_{n=1}^\infty\frac{1}{p_n}=e$ in passing, I just might believe it. – R. Burton Nov 18 '19 at 22:47

My favourite example are Borwein integrals. Consider following expressions \begin{aligned} \frac{2}{\pi} \int_0^\infty \frac{\sin(x)}{x} dx &= 1\\ \frac{2}{\pi} \int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3} dx &= 1\\ \frac{2}{\pi} \int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\frac{\sin(x/5)}{x/5} dx &= 1\\ \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots&\dots\dots\\ \frac{2}{\pi} \int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\cdots\frac{\sin(x/13)}{x/13} dx &= 1 \end{aligned} i.e. you have sequence $$1,1,1,1,1,1,1$$. Would you expect the next number to be $$1$$? Well $$\frac{2}{\pi} \int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\cdots\frac{\sin(x/15)}{x/15} dx = \frac{467807924713440738696537864469}{467807924720320453655260875000}.$$
Seven ones is too little to make conclusions? Than what about a slight modification: $$\frac{2}{\pi} \int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/101)}{x/101}\cdots\frac{\sin(x/(100n+1))}{x/(100n+1)} dx = 1$$ for all $$n < e^{99} \approx 10^{43}$$ and than the pattern breaks.
It is kind of non-obvious that the harmonic sum ($$\sum_{i=1}^\infty 1/i$$) does not converge when you just look at the partial sums. An example that is a bit different (and maybe not what was asked for) is the polynomial $$n^2+n+41$$. It has prime values for all integral $$n$$ with $$0\leq n \leq 39$$.