**Lemma** If $a_{n} \geq 0$ and $\sum_{n=1}^{\infty}< \infty$ then every [[Rearrangement of Series|rearrangement]] has the same limit. **Proof** Let $b_{n}$ be a rearrangement, so that $b_{n} = a_{\sigma(n)}$ For each $N$, let $M_{N} = \max\{\sigma(n): n = 1,2,\dots N\}$ Then $\sum_{n=1}^{N} b_{n}\leq \sum_{n=1}^{M_{N}} a_{n} \leq \sum_{n=1}^{\infty} a_{n}.$So $\sum b_{n} <\infty$ since [[Series with Non-Negative Terms Converges Iff Partial Sums Are Bounded Above|partial sums are increasing and bounded above]] by $\sum a_{n}$ and $\sum b_{n} \leq \sum a_{n}$ by [[Limits of Real Sequence Preserve Weak Inequalities]]. If we let $L_{N} = \max\{\sigma^{-1} (r): r= 1,\dots,N \}$ then the first $N$ terms of $(a_{n})$ are included in the first $L_{N}$ terms of $(b_{n})$. So $\sum_{j=1}^{N} a_{j} \leq \sum_{j=1}^{L_{N}} b_{j}$. So similarly $\sum a_{n} \leq \sum b_{n}$ hence $\sum a_{n} = \sum b_{n}$. **Remark** See Corollary [[If a series is absolutely convergent then every rearrangement has the same limit]].