# Proof: show that this series is absolutely convergent using the given hint

I've posted the question yesterday: see Proof: show that this series converges absolutely to an arbitrary real number. I'm having some difficulties applying Jack D'Aurizio's hint. The problem states the following: Suppose that $\sum a_n$ is a conditionally convergent series. Then for every $\beta \in \mathbb{R}$ there exists a partition $\mathbb{N}=\bigcup_{i=1}^{\infty}B_i$, where each set $B_i$ is finite, and the series $\sum_i(\sum_{n \in B_i}a_n)$ converges absolutely to $\beta$.

Follow Jack D'Aurizio's construction and abbreviate $$b_i:= \sum_{n\in B_i}a_n=\sum_{j=N(i-1)+1}^{N(i)}a_{\sigma(i)}.$$ we have for $m\in\mathbb N$, $$\sum_{i=1}^m b_i=\sum_{i=1}^m \sum_{n\in B_i}a_n=\sum_{j=0}^{N(m)}a_{\sigma(j)}.$$ By definition of $N(m)$, for any $N>N(m)$, $$\left|\beta-\sum_{j=0}^{N}a_{\sigma(j)}\right|< \frac1{2^m}.$$ Consequently, $$\left|\beta-\sum_{i=1}^nb_i\right|<\frac1{2^m}$$ whenever $n>m$. This shows at least that we have convergence, $$\sum_{i=1}^m \sum_{n\in B_i}a_n=\beta.$$ To see that the convergence is absolute, observe that $|b_n|<2\cdot \frac1{2^m}$ for $n>m$, so that the series is dominated by a geometric series.