# Can this sum ever converge?

If I have a strictly increasing sequence of positive integers, $n_1<n_2<\cdots$, can the following sum converge?

$$\sum_{i=1}^\infty \frac{1}{n_i} (n_{i+1}-n_{i})$$

I suspect (and would like to prove) that it always diverges. Haven't made much progress so far, though.

On a related note, is there any characterization of which subsequences of $1/n$ have a convergent sum?

• Heuristic argument: if $f$ is $C^1$ increasing and tends to $+\infty$, then $\int^{+\infty} \frac{f'}{f}$ diverges as an antiderivative is $\log f$. Morality: use integral comparison. – Julien May 24 '13 at 18:44
• $$\frac{n_{i+1}-n_{i}}{n_i} \ge \log\frac{n_{i+1}}{n_i} \implies \sum_{i=1}^N\frac{n_{i+1}-n_{i}}{n_i} \ge \log\frac{n_{N+1}}{n_1} \to \infty \text{ as } N \to \infty$$ – achille hui May 24 '13 at 18:48
• ^ Let $n_i=2-\frac{1}{i}$, this is a strictly increasing set of positive integers, and $\log\frac{n_{N+1}}{n_1}\to\log 2$ as $N\to\infty$. – JLA May 25 '13 at 21:22
• @JLA I propose we call $2-\frac{1}{57}=\frac{113}{57}$ JLA's integer, and maybe, if you're lucky, it will end up next to Grothendieck's prime 57 in history books. – Julien May 25 '13 at 23:36
• Haha, wow, my mistake. – JLA May 30 '13 at 9:30

To expand on my comment (oops, and even more on Achille Hui's comment which appeared when I was typing): $$\frac{n_{i+1}-n_i}{n_i}\geq \int_{n_i}^{n_{i+1}}\frac{1}{t}dt\;\Rightarrow \; \sum_{i=1}^K\frac{n_{i+1}-n_i}{n_i}\geq \int_{n_1}^{n_{K+1}}\frac{1}{t}dt=\log\left(\frac{n_{K+1}}{n_1}\right)\rightarrow +\infty.$$ Of course, $n_K$ tends to $+\infty$ as $n_k\geq K$ for every $K$ by induction.
For your other question, a partial answer can be given with the help of the notion of natural/asymptotic density. If $\liminf \frac{k}{n_k}>0$, then there exists $m>0$ and $K$ such that $\frac{k}{n_k}\geq m$ for every $k\geq K$, whence $\sum_{k\geq K}\frac{1}{n_k}\geq m\sum_{k\geq K}\frac{1}{k}=+\infty$. That is, if the set $\{n_k\;;\;k\geq 1\}$ has positive asymptotic lower density, then the series $\sum_{k\geq 1}\frac{1}{n_k}$ diverges. But the converse is false. For instance, the set of prime numbers $\mathcal P$ has null asymptotic density and yet $\sum_{p\in\mathcal{P}}\frac{1}{p}=+\infty$.
• Thanks! I had got stuck here while trying to prove that if $x_n$ is a positive, decreasing sequence, and $\sum x_n$ converges, then $nx_n\rightarrow 0$. This will directly imply that $\{n_k,k\geq 1\}$ should have zero asymptotic density. Are there any sufficient conditions for convergence? – user56914 May 25 '13 at 13:32
• Except for obvious ones such as, for instance, $n_k=O(1/k^p)$ for some $p>1$ (which has nothing to do with the fact that $n_k$ are integers), I don't know. – Julien May 25 '13 at 13:37
• @Amudhan Oops. Of course, I meant $1/n_k = O(1/k^p)$ for some $p>1$. – Julien May 25 '13 at 15:11