# Rearrangement of divergent series with positive terms.

I am given a question: Can a series with non-negative terms be turned via rearrangement into series with different sum? My solution: No:

If $$\sum a_n$$ is a series with non-negative terms, then the sequence of partial sums given by $$s_n=\sum_{k=1}^n$$ is monotone increasing, thus has a limit in $$\mathbb{R} \cup \{ \infty \}$$. So, if $$\sum a_n$$ converges, then $$\sum|a_n|$$ converges too, because $$a_n\geq 0$$ as assumed. From Riemann rearangement theorem any of its reordering converges and has the same sum. I am stuck with the part where $$\sum a_n$$ diverges to $$\infty$$. I wish to show that for any bijection $$\varphi:\mathbb{N}\rightarrow \mathbb{N}$$ the series $$\sum a_{\varphi(n)}$$ diverges to $$\infty$$. I tried looking at $$s'_n=\sum_{k=0}^n a_{\varphi(k)}$$ and show that this diverges. Isn't this somehow related to the sequence or the subsequence of $$s_n$$? How to proceed?

Let $$\sum_{m=1}^\infty a_{\phi(m)}\le K$$ for some $$K>0$$. Consider a partial sum of $$\sum_{n=1}^Na_n$$. Then for $$n=1,\ldots,N$$ there exists $$m_n$$ such that $$a_n = a_{\phi(m_n)}$$. Hence, setting $$M = \max\{m_n : n=1,\ldots,N\}$$, $$s_N = \sum_{n=1}^Na_n = \sum_{n=1}^Na_{\phi(m_n)}\le\sum_{m=1}^{M}a_{\phi(m)}\le\sum_{m=1}^\infty a_{\phi(m)}\le K.$$ This holds for every $$N$$ and hence $$(s_N)$$ is increasing and bounded, thus convergent.