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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.

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Any rearrangement of an absolutely convergent series like yours converges, and to the same sum.

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 $\frac{1}{2}, \frac{1}{2^2}, \dots,\frac{1}{2^n}$.

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  • $\begingroup$ 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. $\endgroup$
    – vadim123
    Apr 23 '13 at 16:02

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