I wonder if given any conditionally convergent series $S=\sum_{k=1}^{\infty}a_k$ we can do the estimation related to the triangular inequality and $\textit{reordering}$ the series, i.e., \begin{align} \left|S=\sum_{k=1}^{\infty}a_k\right|\leq\left|\sum_{k=1}^{\infty}a_{2k}+\sum_{k=1}^{\infty}a_{2k+1}\right|\leq\left|\sum_{k=1}^{\infty}a_{2k}\right|+\left|\sum_{k=1}^{\infty}a_{2k+1}\right|? \end{align} If we assume that $\sum_{k=1}^{\infty}a_{2k}$ &$\sum_{k=1}^{\infty}a_{2k+1}$ are both well defined and can be eual to $\infty$ does the triangular inequality hold?

Also, is there any good reference that we can know more about conditionally convergent series like how to know that how much cancellation of each term we can get or when we can know more about periodicity of "sign" like harmonic / alternating series ?

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    $\begingroup$ The expression $\sum_{k=1}^{\infty} a_{2k} + \sum_{k=1}^{\infty} a_{2k+1}$ may not be defined due to $\infty - \infty$ issues, such as when $a_k=(-1)^k/k$. Also, what is meant by your "..." at the end ? $\endgroup$ – Michael Oct 23 '18 at 1:52
  • $\begingroup$ @Michael I fixed it. Thank you. If the both are infinity then we still can bound $S$ by infinity right? $\endgroup$ – Thomas Oct 23 '18 at 1:57

Note that you need to re-index so that you include $a_1$.


Suppose $\sum_{n=1}^{\infty} a_n$, $\sum_{k=1}^{\infty} a_{2k}$, and $\sum_{k=1}^{\infty} a_{2k-1}$ are all well defined (possibly infinite) and at most one of the two sums $\sum_{k=1}^{\infty} a_{2k}$, and $\sum_{k=1}^{\infty} a_{2k-1}$ has an infinite value (either $\infty$ or $-\infty$). Then $$ \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} a_{2k} + \sum_{k=1}^{\infty}a_{2k-1}$$ and $$ \left|\sum_{n=1}^{\infty} a_n\right| \leq \left|\sum_{n=1}^{\infty} a_{2k}\right| + \left|\sum_{k=1}^{\infty}a_{2k-1}\right|$$

Proof: For each positive integer $M$ we have $$ \sum_{n=1}^{2M} a_{n} = \sum_{k=1}^{M}a_{2k} + \sum_{k=1}^{M}a_{2k-1} \quad (Eq. *)$$ Taking limits as $M\rightarrow\infty$ and noting that the right-hand-side avoids the undefined case of $\infty-\infty$ gives the first equality. On the other hand, taking absolute values of both sides of Eq. (*) and using the triangle inequality gives $$ \left|\sum_{n=1}^{2M} a_{n}\right| \leq \left|\sum_{k=1}^{M}a_{2k}\right| + \left|\sum_{k=1}^{M}a_{2k-1}\right|$$ Taking limits as $M\rightarrow\infty$ completes the proof. $\Box$


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