A LIMIT COMPARISON THEOREM FOR REARRANGEMENTS

In the previous section we considered the convergence behavior of some special series. This section is devoted to a more general convergence result. Similar to the rich theory of the convergence of positive series, we will prove a comparison theorem. This will allow us to study the convergence of rearrangements of series with more complicated terms.

Theorem 2   Let

$$\sum_{n=1}^{\infty} a_n(-1)^{n-1}\qquad\mbox{and}\qquad \sum_{n=1}^{\infty} b_n(-1)^{n-1}$$

be two conditionally convergent series, which satisfy the assumptions spelled out in the introduction of this paper. Assume that there is a positive constant $C$ such that.

$$\lim_{n\to\infty}\frac{a_n}{b_n}=C.$$

Then any rearrangement of

$$\sum_{n=1}^{\infty}a_n(-1)^{n-1}$$

will converge if and only if the corresponding rearrangement of

$$\sum_{n=1}^{\infty}b_n(-1)^{n-1}$$

converges.

Proof: To facilitate the proof of this theorem we need to introduce some notation. Consider a given rearrangement of the alternating series

$$\sum_{n=1}^{\infty}a_n(-1)^{n-1}$$

with associated sequence s $p_n$ and $q_n$ of positive and negative terms. Without loss of generality we assume that $p_n>q_n$ for sufficiently large $n$. Let

$$S_N(a)$$

denote the $N$-th partial sum of this rearrangement, and

$$S_N(b)$$

denote the $N$-th partial sum of the same rearrangement of

$$\sum_{n=1}^{\infty}b_n(-1)^{n-1}.$$

Moreover, let $T_N(a)$ and $T_N(b)$ denote the partial sums of the corresponding alternating series. Furthermore, let

$$\alpha:[1,\infty)\to [0,\infty) \qquad\mbox{and}\qquad \beta:[1,\infty)\to[0, \infty)$$

be two continuous functions with anti derivatives $A$ and $B$ such that

$$\alpha(n)=a_n\qquad\mbox{and} \qquad \beta(n)=b_n.$$

Finally, let

$$B(x)=\int_1^x \beta(t)\,dt, \qquad\mbox{and}\qquad A(x)= \int_1^x \alpha(t)\,dt$$

Let $\epsilon>0$ then there exists an $N_0$ such that

$$b_n(C-\epsilon)<a_n<b_n(C+\epsilon)$$

for all $n\ge \frac{q_N}2+1$. Let $N\ge N_0$, then

$$\begin{eqnarray*} S_N(a)&=&T_{q_N}(a)+\sum_{n=q_N+1}^{p_N} a_{2n-1}\\ &\le&T_{q_N}(a)+(C+\epsilon)\sum_{n=q_N+1}^{p_N} b_{2n-1} \end{eqnarray*}$$

Analogously, we get

$$\begin{eqnarray*} S_M(a)&=&T_{q_M}(a)+\sum_{n=q_M+1}^{p_M} a_{2n-1}\\ &\ge&T_{q_M}(a)+(C-\epsilon)\sum_{n=q_M+1}^{p_M} b_{2n-1} \end{eqnarray*}$$

for $M\ge N_0$. Subtracting the second inequality from the first we get

$$\begin{eqnarray*} S_N(a)-S_M(a) &\le& T_{q_N}(a)-T_{q_M}(a) +C\sum_{n=q_N+1}^{p... ...n=q_N+1}^{p_N} b_{2n-1}+ \epsilon\sum_{n=q_M+1}^{p_M} b_{2n-1} \end{eqnarray*}$$

In this step we used the decomposition of $S_N(b)$ into $T_{q_M}(b)$ and a positive remainder term, and the fact that $x\le \vert x\vert$. On the right hand side of the last inequality we may interchange $M$ and $N$ without changing the value of the right hand side. This implies that the same inequality applies to

$$S_M(a)-S_N(a)$$

and therefore

$$\begin{eqnarray*} \left\vert S_N(a)-S_M(a)\right\vert&\le&\left\vert T_{q_N}(a)... ...n=q_N+1}^{p_N} b_{2n-1}+ \epsilon\sum_{n=q_M+1}^{p_M} b_{2n-1} \end{eqnarray*}$$

Next, from the proof of Theorem 1 we have that

$$\sum_{n=q_N+1}^{p_N} b_{2n-1}\le \frac12\int_{2q_N}^{2p_N} \beta(t)\,dt+b_{2q_N+1}=B(2p_N)-B(2q_N)+b_{2q_N+1}.$$

Now assume that $S_N(b)$ converges, then by Theorem 1 $B(2p_N)-B(2q_N)$ converges to a finite number and hence there exists a $K>0$ such that

$$B(2p_N)-B(2q_N)+b_{2q_N+1}<K$$

for all $N$. Next, since $S_N(b)$, $T_N(a)$, and $T_N(b)$ all converge, they are Cauchy sequences and there exists a $M_0$ such that

$$\begin{eqnarray*} \left\vert T_{q_N}(a)-T_{q_M}(a)\right\vert<\epsilon\\ \lef... ...\vert<\epsilon\\ \left\vert S_N(b)-S_M(b)\right\vert<\epsilon \end{eqnarray*}$$

Hence, for $M,N\ge \max\left\{N_0,M_0\right\}$

$$\left\vert S_N(a)-S_M(a)\right\vert<\epsilon+2C\epsilon+2K\epsilon$$

and therefore it is a Cauchy sequence and it converges. The opposite direction is proved completely analogously.

$\square$

We illuminate the use of Theorem 2 by an example. Consider the series

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{\ln\sin\frac1n}$$

The reader can easily verify that

$$a_n=\frac{-1}{\ln\sin\frac1n}$$

satisfies the conditions of both theorems. Now using l'Hospital's rule we get

$$\lim_{x\to\infty}\frac{-\ln\sin\frac1x}{\ln x}=1.$$

Therefore, by Corollary 6, any rearrangement of this series converges if and only if

$$\lim_{n\to\infty}\frac{p_n}{q_n}=1.$$

Remarks:

1. To best of our knowledge, Theorem 2 is, at least in the form given, a new result, albeit not very useful. A. Pringsheim [7] does a comparison of conditionally convergent series with the alternating harmonic series, by comparing whether
   
   $$\lim_{n\to\infty} na_n$$
   
   
   is finite or not. This result follows from Theorem 2 by using the alternating harmonic series as one of the two series used in Theorem 2.
2. The condition that
   
   $$0<\lim_{n\to\infty}\frac{a_n}{b_n}<\infty$$
   
   
   is only a sufficient condition for rearrangements of the two series two converge together. As we have seen in Corollaries 2 and 3 rearrangements of,
   
   $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\quad\mbox{and} \quad \sum_{n=2}^{\infty}\frac{\log n}{n}(-1)^n$$
   
   
   converge together, but the condition of Theorem 2 is obviously not satisfied.
3. The entire paper only applies to series with eventually decreasing terms. There are, however series which do not satisfy this property and which are still conditionally convergent and subject to Riemann's Theorem.