Why this is O(1) (the big oh)??

Question

How can I prove that this sum

$$\sum_{n=2}^{\infty}\frac{\log(n)}{n(n-1)}=O(1) $$

that means that is bounded or is convergent? how can i prove it?

Answer 1

This is a non-sensical usage of $O(\cdot)$. But it could mean that

$$\sum_{n=2}^{N}\frac{\log(n)}{n(n-1)}=O(1)\tag{1}$$

Big $O$ is about a function, and the left side of your expression is not a function of any variable, it is either a value or undefined.

Since all summands here are positive, (1) is equivalent to saying the series converges.

To prove this is convergent, you might note that:

$$\frac{\log n}{n(n-1)}=\frac{\log(n)}{n-1}-\frac{\log(n)}{n}$$

So:

$$\sum_{n=2}^{N}\frac{\log(n)}{n(n-1)}=\frac{\log 2}{1}-\frac{\log(N)}{N}+\sum_{n=2}^{N-1} \frac{\log(n+1)-\log(n)}{n}$$

But $$0<\log(n+1)-\log(n)=\log\left(1+\frac{1}{n}\right)<\frac{1}n$$

So $0<\frac{\log(n+1)-\log(n)}{n}<\frac{1}{n^2}$, and since $\sum_{n=2}^{\infty}\frac{1}{n^2}$ converges, and $\log(2)-\frac{\log(N)}{N}=O(1)$, you get your result.

Answer 2

By comparison with the $p$-series $\sum_n \frac{1}{n^{3/2}}$ (for instance -- any constant $1<p<2$ will work here), we have that the series you are considering converges.

Thus, it is both convergent and bounded.

However, writing $O(1)$ here is (at least as far as I can tell) a strange notation, and an abuse of the asymptotic $O(\cdot)$ notation: there is not asymptotics at play here (no variable going to infinity; no free variable at all, for that matter.)

Answer 3

In general, $O(1)$ means it's bounded. Since in this case the terms are positive, that implies it is convergent. You can use Limit Comparison Test with $\sum n^{-p}$ where $1 < p < 2$.