Convergence of Rearranged Series of Real Numbers

Convergence of Rearranged Series of Real Numbers

Recall from the Rearrangements of Terms in Series of Real Numbers page that if $(a_n)_{n=1}^{\infty}$ is a sequence of real numbers, $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is the corresponding series, and $f : \mathbb{N} \to \mathbb{N}$ is a bijection, then if we define $b_n = a_{f(n)}$ then the new series $\displaystyle{\sum_{n=1}^{\infty} b_n}$ is said to be a rearrangement of $\displaystyle{\sum_{n=1}^{\infty} a_n}$.

We previously alluded that a rearrangement of the terms in a series may converge to a different sum than the original series. Under certain conditions though, we can preserve the sum of a series under a rearrangement. We first make a couple of quick definitions which we will look at more in-depth later.

Definition: A series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is said to be Absolutely Convergent if $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ converges. A series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ that converges but $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ does not converge is said to be Conditionally Convergent.

We are now ready to state under what condition the sum of a series will be preserved under rearrangements.

Theorem 1: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be an absolutely convergent series that converges to the sum $A$. Then any rearrangement of terms, $\displaystyle{\sum_{n=1}^{\infty} b_n}$ is absolutely convergent and will also converge to $A$.
  • Proof: Let $f : \mathbb{N} \to \mathbb{N}$ and define $b_n = a_{f(n)}$ for all $n \in \mathbb{N}$. We begin by showing that the rearrangement series $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges absolutely.
  • Since $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges absolutely, we have that $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid = A}$ for some $A \in \mathbb{R}$.
  • Let $(B_n^*)_{n=1}^{\infty}$ denote the sequence of partial sums to the series $\displaystyle{\sum_{n=1}^{\infty} \mid b_n \mid}$. Notice that $B_n^* \leq B_{n+1}^*$ for all $n \in \mathbb{N}$, so this sequence of partial sums is increasing. Moreover, for all $n \in \mathbb{N}$ we have that:
\begin{align} \quad B_n^* = \sum_{k=1}^n \mid b_k \mid = \sum_{k=1}^{n} \mid a_{f(k)} \mid \leq \sum_{k=1}^{\infty} \mid a_k \mid = A \end{align}
  • So $(B_n^*)_{n=1}^{\infty}$ is an increasing sequence that is bounded above. So $(B_n^*)_{n=1}^{\infty}$ converges which shows that $\displaystyle{\sum_{n=1}^{\infty} \mid b_n \mid}$ converges. So $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges absolutely.
  • Let $\epsilon > 0$. We now wish to establish that $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges to $A$. Let $(A_n)_{n=1}^{\infty}$ be the sequence of partial sums to the series $\displaystyle{\sum_{n=1}^{\infty} a_n}$. Since $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges to $A$ we have that $\displaystyle{\lim_{n \to \infty} A_n = A}$, so, for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $n \geq N_1$ then:
\begin{align} \quad \mid A_n - A \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}
  • Furthermore, there exists an $N_2 \in \mathbb{N}$ such that if $n \geq N_2$ then:
\begin{align} \quad \sum_{k=N+1}^{\infty} \mid a_k \mid < \frac{\epsilon}{2} \quad (**) \end{align}
  • Let $N' = \max \{ N_1, N_2 \}$, consider the set $\{ 1, 2, ..., N' \}$. Then there exists an $N \in \mathbb{N}$, $N' \leq N$, such that:
\begin{align} \quad \{1, 2, ..., N' \} \subseteq \{ f(1), f(2), ..., f(N) \} \end{align}
  • Hence:
\begin{align} \quad \{a_1, a_2, ..., a_{N'} \} \subseteq \{ a_{f(1)}, a_{f(2)}, ..., a_{f(N)} \} \\ \quad \{a_1, a_2, ..., a_{N'} \} \subseteq \{ b_1, b_2, ..., b_{N} \} \end{align}
  • So for $n \geq N$, we have that $n \geq N_2$ and so $(**)$ holds. So:
\begin{align} \quad \mid B_n - A_{N} \mid &= \mid (b_1 + b_2 + ... + b_n) - (a_1 + a_2 + ... + a_{N}) \mid \\ \quad &= \mid (a_{f(1)} + a_{f(2)} + ... + a_{f(n)}) - (a_1 + a_2 + ... + a_{N}) \mid \\ \quad &= \mid a_{N+1} + a_{N+2} + ... + a_{f(n)} \mid \\ \quad & \leq \mid a_{N+1} \mid + \mid a_{N+2} \mid + ... + \mid a_{f(n)} \mid \\ \quad & \leq \sum_{k=N+1}^{f(n)} \mid a_k \mid \\ \quad & \leq \sum_{k=N+1}^{\infty} \mid a_k \mid \\ \quad & < \frac{\epsilon}{2} \quad (***) \end{align}
  • Hence, for $n \geq N$ both $(*)$ and $(***)$ hold, and so:
\begin{align} \quad \mid B_n - A \mid = \mid B_n - A_N + A_N - A \mid \leq \mid B_n - A_N \mid + \mid A_N - A \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
  • So for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\mid B_n - A \mid < \epsilon$. So $\displaystyle{\lim_{n \to \infty} B_n = A}$ which shows that $\displaystyle{\sum_{n=1}^{\infty} b_n = A}$. $\blacksquare$

So any rearrangement of terms in an absolutely convergent series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ will converge to the same sum as $\displaystyle{\sum_{n=1}^{\infty} a_n}$.

We now state (but not prove) a theorem which says that if $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is only conditionally convergent, then the same cannot be said.

Theorem 2: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be a conditionally convergent series that converges to the sum $A$. Then for any $X$ and $Y$ such that $- \infty \leq Y \leq X \leq \infty$ there exists a rearrangement $\displaystyle{\sum_{n=1}^{\infty} b_n}$ such that if $(B_n)_{n=1}^{\infty}$ denotes the sequence of partial sums to this rearrangement, then $\displaystyle{\limsup_{n \to \infty} B_n = X}$ and $\displaystyle{\liminf_{n \to \infty} b_n = Y}$.

Note that if we set $X = Y = c$, then such an arrangement $\displaystyle{\sum_{n=1}^{\infty} b_n}$ will converge to $c$. Hence we can rearrange the term of a conditionally convergent series to converge to any real number that we so choose.

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