<s>[INST] Infinite sums - order of terms say we had an infinite sum $ \displaystyle \sum_{n=0}^\infty a_n $ where $ a_n  = \dfrac{1}{2}  +1 + \dfrac{1}{8} + \dfrac{1}{4} + \dfrac{1}{32} + ... $ this is obviously a rearrangment of $\displaystyle \sum_{n=0}^\infty \dfrac{1}{2^n} $ but does the order of the sum matter? I know by the general theorem that as $\displaystyle \sum_{n=0}^\infty \dfrac{1}{2^n} $ converges absolutely then any rearrangement also converges to the same sum - but how would I go to prove that $a_n$ converges WITHOUT using that theorem? e.g. using the root test
also, why is it for some sums - the order matters, while others it does not
 [/INST] If a series involves only positive (or at least non-negative) terms, then any rearrangement of the series has the same limit in the sum. One way to prove this is to note that if $a_n \ge 0$ for all $n$, then
$$\sum\limits_{n = 0}^{\infty} a_n = \sup_{A \text{ finite}} \sum_{n \in A} a_n$$

To prove that this series converges directly, however, note that we can just compute 
\begin{align*}
1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{4} + \frac{1}{32} + \frac{1}{16} + \dots &= \frac{2}{2} + \frac{1}{2} + \frac{1}{8} + \frac{2}{8} + \frac{2}{32} + \frac{1}{32} + \dots \\
&= \frac{3}{2} + \frac{3}{8} + \frac{3}{32} + \dots \\
&= \frac{3}{2} \left(1 + \frac{1}{4} + \frac{1}{16} + \dots \right) \\
&= \frac{3}{2} \frac{1}{1 - \frac{1}{4}} \\
&= 2
\end{align*}
where we've used a geometric series.
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