Can $\sum_n \frac{1}{x_{n_{ik}}}$ still diverge if $\sum_n \frac{1}{x_{n_{i}}}$ diverge and $x_n$ is nondecreasing?

Question

Let $(x_n)_{n\in\Bbb N}$ be a nondecreasing sequence of positive real numbers and suppose that there exists a sequence $(n_i)_{i\in\Bbb N}$ such that $$\sum_{i=1}^\infty \frac{1}{x_{n_i}} = +\infty.$$ Can I conclude that, for all $k$ big, the series $$\sum_{i=1}^\infty \frac{1}{x_{n_{ki}}}$$ diverges as well or there are counterexamples?

It seems counter-intuitive to me that we still have $\sum_{i=1}^\infty 1/{x_{n_{ki}}}$ even if $k$ is big, but I couldn't find a counterexample so I guess it can be somehow proved.

Answer 1

Yes, it diverges as well: here's a proof. Lets write $y_i = x_{n_i}$ because there are too many subscripts and the first subsequence $n_i$ is not important at all.

Since $y_i$ is non-decreasing, for the $k$ indices $j=ki+1,\dots,k(i+1)$, we have $$ y_{ki} \le y_j \iff \frac1{y_j} \le \frac1{y_{ki}}$$ which means that, since $k$ is fixed, $$\sum_{j=ki+1}^{k(i+1)} \frac1{y_j} \le \frac{k}{y_{ki}} $$ Taking a sum from $i=1$ to $N$,
$$ \sum_{i=1}^{N-1}\frac k{y_{ki}} \ge \sum_{i=1}^N \sum_{j=ki+1}^{k(i+1)} \frac1{y_j} = \sum_{\iota=k+1}^{k(N+1)}\frac1{y_\iota} \xrightarrow[N\to\infty]{} \infty .$$

Therefore $ \sum_{i=1}^{\infty}\frac 1{y_{ki}}=\infty$ as well.

If instead of $i$, you put a function of $i$ (e.g. $i^2$) then something else could happen.

Answer 2

An obvious answer is if $x_n$'s are bounded above (non-decreasing, or even increasing does not guarantee unboundedness), say $x_n\leq \beta$, then for any infinite subset $S\subseteq\mathbb{N}$, $\sum_{n\in S} \frac{1}{x_n} \to\infty$.

Now if we assume that $x_n$'s are unbounded, taking $x_n=n$ will give you an example for your question.