# Rearrangement of series

The series $$\frac{1}{2^2}+\frac{1}{2}+\frac{1}{2^4}+\frac{1}{2^3}+...$$ is a rearrangement of the geometric series $$\sum_{n=1}^\infty\frac{1}{2^n}$$

Does it converge or converge absolutely?

(Theorem: If the series $\sum_{n=1}^\infty a_n$ converges, then any series obtained by grouping the series of $\sum_{n=1}^\infty a_n$ is also convergent and has the same value as $\sum_{n=1}^\infty a_n$)

My confusion: Based on the theorem, the series $$\frac{1}{2^2}+\frac{1}{2}+\frac{1}{2^4}+\frac{1}{2^3}+...$$ is convergent since $\sum_{n=1}^\infty\frac{1}{2^n}$ is a geometric series with |r|<1, but if i work from the series $\frac{1}{2^2}+\frac{1}{2}+\frac{1}{2^4}+\frac{1}{2^3}+...$, my answer is not convergent. And furthermore, i found out that limsup of this series = 2 $\not\lt$ 1 i.e. the series do not converge absolutely.

To see convergence in your case, note that the partial sums are increasing and bounded by $1$. And $1$ is the least upper bound for the partial sums. This is because for any $n$ there is an $N$ such that the rearranged terms up to the $N$-th include all of $\frac{1}{2}, \frac{1}{2^2}, \dots,\frac{1}{2^n}$.
• In this problem, rearrangement is not necessary. Grouping the terms pairwise gives the geometric and absolutely convergent series $\frac{3}{2^2}+\frac{3}{2^4}+\frac{3}{2^6}+\cdots$. Breaking these pairs in in the other way gives the original geometric series. Apr 23 '13 at 16:02