0
$\begingroup$

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 $\beta$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.