Riemann’s theorem on rearrangements

If the map $n\mapsto n^{\prime }$ is a bijection on $\mathbb{N}$, we say that the sequence $(a_{n^{\prime }})$ is a rearrangement of $(a_n)$.

The following theorem, which is due to Riemann, shows that the convergence of a conditionally convergent series depends so much on the order of its terms; in particular, a conditionally convergent series can be made to converge to any real number by changing the order of its terms.

Theorem (Riemann series theorem).Let $(a_n)$ be a sequence in $ℝ$ such that ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n$ converges but ${\sum }_{n=1}^{\mathrm{\infty }}|a_n|=\mathrm{\infty }$, i.e, $\sum a_n$ is conditionally convergent. Let be arbitrary. Then there exists a rearrangement $(a_{n^{\prime }})$ such that

$$\underset{N}{lim\; inf}\sum _{n^{\prime }=1}^N{a}_{n^{\prime }}=\alpha \mathit{\hspace{1em}\hspace{1em} }\text{and}\mathit{\hspace{1em}\hspace{1em} }\underset{N}{lim\; sup}\sum _{n^{\prime }=1}^N{a}_{n^{\prime }}=\beta .$$

Proof.Let $a_n^{+}=\max \{0,a_n\}$ and $a_n^{-}=\min \{0,-a_n\}$. Then we have $a_n=a_n^{+}-a_n^{-}$ and $|a_n|=a_n^{+}+a_n^{-}$. Since , both $\sum a_n^{+}$ and $\sum a_n^{-}$ diverge or converge simultaneously. But since $\sum |a_n|=\mathrm{\infty }$, we see that at least one of $\sum a_n^{+}$ and $\sum a_n^{-}$ must diverge. It follows that $\sum a_n^{+}=\mathrm{\infty }$ and $\sum a_n^{-}=\mathrm{\infty }$.

Also by the $n$th term test, ${lim}_n{a}_n^{+}={lim}_n{a}_n^{-}=0$.

Now we pass to subsequence of $(a_n^{+})$ by removing all terms with $a_n^{+}=0$ and $a_n\ne 0$. For $a_n^{-}$, we remove all terms with $a_n^{-}=0$. Let us denote the subsequences still as $(a_n^{+})$ and $(a_n^{-})$. Since only zeros have been removed, $\sum a_n^{+}$ and $\sum a_n^{-}$ are still divergent.

Now we will define integers $m_j$ and $k_j$ for $j\in \mathbb{N}$, and consider the series

$a_1^{+}$$+a_2^{+}+\mathrm{\cdots }+a_{m_1}^{+}$

$-a_1^{-}-a_2^{-}-\mathrm{\cdots }-a_{k_2}^{-}$

$+a_{m_1+1}^{+}+a_2^{+}+\mathrm{\cdots }+a_{m_2}^{+}$

$-a_{k_1+1}^{-}-a_2^{-}-\mathrm{\cdots }-a_{k_2}^{-}+$

$\mathrm{ }\mathit{\hspace{1em}}\mathrm{⋮}$

This series is clearly a rearrangement of $\sum a_n$, by our choice of the subsequences $a_n^{+}$ and $a_n^{-}$.

We pick up two sequences ${\alpha }_j$ and ${\beta }_j$ such that ${\alpha }_j\to \alpha$ , ${\beta }_j\to \beta$, ${\alpha }_n\le {\beta }_n$ and ${\beta }_1>0$. We choose $m_1$ such that ${\sum }_{n=1}^{m_1}\ge {\beta }_1$ but . We choose $k_1$ such that ${\sum }_{n=1}^{m_1}-{\sum }_{n=1}^{k_1}\le {\alpha }_1$ but ${\sum }_{n=1}^{m_1}-{\sum }_{n=1}^{k_1-1}>{\alpha }_1$. We continue this way, inductively.

Since ${lim}_n{a}_n^{+}={lim}_n{a}_n^{-}=0$, the subsequences of the sequence of partial sums that end with $a_{m_j}^{+}$ and $a_{k_j}^{-}$ converge to $\beta$ and $\alpha$, and it can be seen that no subsequence can be found with a limit larger than $\beta$ or lower than $\alpha$. $\mathrm{\square }$