Let $1=n_0Let $1=n_0<n_1<n_2<\ldots$ be an increasing sequence of positive integers. Is it true that $\displaystyle\sum_{i=0}^{\infty} \frac{n_{i+1}-n_i}{n_{i+1}} $ diverges to $+\infty?$
For example, if $n_1=2,n_2=3,n_3=53,n_4=54,$ then the first few terms are $\frac{1}{2}, \frac{1}{3}, \frac{50}{53}, \frac{1}{54}.$ Since $\frac{50}{53}<\displaystyle\sum_{k=4}^9 \frac{1}{k},$ this makes me think it might be possible to come up with a counter-exmaple.
 A: The series diverges. Hint: for a fixed $k$ and $\ell$ big enough we have $$\frac{n_{k+1}-n_k}{n_{k+1}}+\frac{n_{k+2}-n_{k+1}}{n_{k+2}}+\ldots+\frac{n_{\ell}-n_{\ell-1}}{n_{\ell}}>\frac{n_{k+1}-n_k}{n_{\ell}}+\frac{n_{k+2}-n_{k+1}}{n_{\ell}}+\ldots+\frac{n_{\ell}-n_{\ell-1}}{n_{\ell}} = \frac{n_\ell-n_k}{n_\ell}\ge \frac 12$$
A: EDIT: I missed the assumption that the sequences has to consist of integers, by bad. Nonetheless I'll leave my answer up regardless as it might still bring some insight to the problem.
Notice first that
$$\sum_{j=1}^\infty\frac{n_{j+1}-n_j}{n_{j+1}}=\sum_{j=1}^\infty\left(1-\frac{n_j}{n_{j+1}}\right).$$
One can thus immediately see that if
$$\lim_{j\to\infty}\frac{n_j}{n_{j+1}}\neq1,$$
then the series diverges, so we have to consider sequences where $\frac{n_j}{n_{j+1}}\to1$ if we wish to find a counterexample. Consider the sequence $\{n_j\}_{j\in\mathbb{Z}^+}$ given by
$$n_j=\frac{2j}{j+1}.$$
Now clearly this sequence is strictly increasing, and $n_0=1$. For this sequence we have that
$$1-\frac{n_j}{n_{j+1}}=1-\frac{2j}{j+1}\cdot\frac{j+2}{2(j+1)}=1-\frac{(j+1)^2-1}{(j+1)^2}=\frac{1}{(j+1)^2}.$$
This means that
$$\sum_{j=1}^\infty\frac{n_{j+1}-n_j}{n_{j+1}}=\sum_{j=1}^\infty\frac{1}{(j+1)^2}=\sum_{j=2}^\infty\frac{1}{j^2},$$
which clearly converges. So the conclusion you can draw is that if $\frac{n_j}{n_{j+1}}\not\to1$ as $j\to\infty$, then it diverges, but otherwise it is inconclusive.
