Rearrangement diverges then original series also diverges?

Assume that $(x_n)_{ngeq 1}$ is a sequence of real numbers and that either

$$ sum_{n=1}^{infty} max{x_n, 0} < infty qquadtext{or}qquad sum_{n=1}^{infty} max{-x_n, 0} < infty $$

holds. (This in particular prevents the series $sum_{n=1}^{infty} x_n$ from converging conditionally.) Then we can prove that any rearrangement $(y_n)_{ngeq1}$ of $(x_n)_{ngeq1}$ satisfies

$$ sum_{n=1}^{infty} x_n = sum_{n=1}^{infty} y_n, $$

regradless of whether they converge or not. (Note that each of these sums always has a value in the extended real number line $[-infty,infty]=mathbb{R}cup{-infty,infty}$, and then the equality holds in $[-infty,infty]$.)

It can be proved by focusing on OP's case (where $x_n$'s are non-negative). Indeed, assuming that $x_n geq n$ and $sigma:mathbb{N}_1tomathbb{N}_1$ is a bijection so that $y_n=x_{sigma(n)}$, then for each given $m geq 1$ there exists $N$ such that ${1,dots,m}subseteq{sigma(1),dots,sigma(N)}$. Then for any $ngeq N$, we have

$$ sum_{k=1}^{m} x_k leq sum_{k=1}^{n} y_{k} leq sum_{k=1}^{infty} x_k. $$

Letting $ntoinfty$, this shows that

$$ sum_{k=1}^{m} x_k leq sum_{k=1}^{infty} y_{k} leq sum_{k=1}^{infty} x_k $$

for any $m geq 1$, then letting $m to infty$ proves the desired equality.

A series of nonnegative numbers converges iff the series converges absolutely. A series converges absolutely iff every rearrangement of that series converges to the same limit.

Since there is a rearrangement that does not converge, the original series cannot converge absolutely. But the terms are nonnegative, so the original series diverges.