EXAMPLES AND CONSEQUENCES

In this section we will use Theorem 1 to investigate the convergence of rearrangements of some prominent alternating series. In all cases we do the explicit proofs for $p_N\ge q_N$ (at least eventually). We mention that the other situation can be handled in the same way. The most prominent -- and most intensively investigated -- conditionally convergent series is the alternating harmonic series, indeed several of the papers cited specialize on this topic [3,5]. As we will later see this series is not a very good model, since it is converging so rapidly, in fact it is almost absolutely convergent. Applying Theorem 1 to that series yields the following result.

Corollary 2   For the alternating harmonic series the statement of Theorem 1 becomes:

$$A-T=\frac12 \lim_{N\to \infty}\ln\frac{p_N}{q_N}$$

Proof: In this case $f(t)=\frac1t$ and $F(t)=\ln t$. Applying Theorem 1 gives

$$2A-2T=\lim_{N\to\infty}\left(\ln2p_N-\ln2q_N\right)=\lim_{N\to\infty}\frac{p_N}{q_N}.$$

$\square$

In particular, this Corollary implies that the limit of a rearrangement of the alternating harmonic series is finite, if and only if $\lim_{N\to\infty}\frac{p_N}{q_N}$ is finite. Other authors do not use the ratio we use in this paper, but rather use the asymptotic density of positive terms in the rearrangement, which is defined as

$$\rho=\lim_{N\to\infty}\frac{p_N}{N}.$$

In this notation, the above corollary implies that a rearrangement of the alternating series converges to a finite limit if and only if $0<\rho<1$. The alternating harmonic series is a relatively rapidly converging alternating series and represents as such a limiting case for conditionally convergent series. Corollary 1 also allows us to compute explicit rearrangements converging to a given number. Since in this case it is known that $T=\ln 2$. For example, to construct a rearrangement which converges to $A=\ln 3$ we must ensure that

$$\frac12 \lim_{N\to\infty}\ln\frac{p_N}{q_N}=\ln3-\ln2=\ln\frac32.$$

This can easily be achieved by taking always 9 positive terms followed by 4 negative terms.

In our next example we will investigate a class of slower converging $p$-series.

Corollary 3   Consider the alternating $p$- series

$$T=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^p}$$

with $0<p<1$. Then a rearrangement of this series converges to a finite limit if and only if

$$\lim_{N\to\infty}\frac{p_N}{q_N}=1,$$

i.e. $\rho=\frac12$.

Proof: In this case $f(t)=t^{-p}$ and hence

$$F(t)=\frac1{1-p}\left(t^{1-p}-1\right).$$

The theorem implies that

$$2A-2T=\lim_{N\to\infty}\frac1{1-p}\left((2p_N)^{1-p}-(2q_N)... ...{1-p}}{1-p}\left(\left(\frac{p_N}{q_N}\right)^{1-p}-1\right)$$ (7)

Since

$$\frac{(2q_N)^{1-p}}{1-p}$$

grows without bound as $N\to \infty$, the limit in (7) can only be finite if

$$\lim_{N\to\infty}\left(\left(\frac{p_N}{q_N}\right)^{1-p}-1\right)=0,$$

which proves our assertion. The statement in terms of the asymptotic density $\rho$ follows immediately.

$\square$

In the case of the alternating harmonic series, convergence of a rearrangement is assured if

$$\lim_{N\to\infty}\frac{p_N}{q_N}$$

is a finite positive number. In Corollary 3, this number must be one. This suggests that there are really at most three classes of conditionally convergent series, Namely series where a rearrangement converges to a finite limit if and only if

$$\lim_{N\to\infty}\frac{p_N}{q_N}=1,$$

and series for which rearrangements converge if and only if this limit is a finite positive number. Finally, there is the possibility of series for which a rearrangement also converges if the limit is either $0$ or if the sequence $\frac{p_N}{q_N}$ is unbounded. We will encounter aan example of such a series later in this section.

We will now extend these results to another prominent set of alternating series.

Corollary 4   Let $p<1$ and consider the series

$$\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^p}(-1)^{n-1}.$$

A rearrangement of this series converges to a finite limit if and only if

$$\lim_{N\to\infty}\frac{p_N}{q_N}$$ (8)

is finite and positive.

Proof: Observe that

$$F(x)=\int_2^x \frac1{t(\ln t)^p}\,dt=\int_{\ln 2}^{\ln x}\frac1{s^p}\,ds.$$

The previous Corollary implies that a rearrangement of

$$\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^p}(-1)^{n-1}$$

converges to a finite limit if and only if

$$\lim_{N\to\infty}\frac{\ln 2p_N}{\ln 2q_N}=1 .$$ (9)

Before continuing we remark that unlike in the previous corollary we can allow $p\le 0$ here, since the original alternating series also converges in this case and the argument of the proof of Corollary 3 still works.

We are left to show the equivalence of (9) and (8). To do this suppose first that

$$\lim_{N\to\infty}\frac{p_N}{q_N}=L$$

for some positive real number $L$. Next choose $\epsilon$ such that $0<\epsilon<L$. Then there exists an $N_0$ such that

$$L-\epsilon < \frac{p_N}{q_N} < L+ \epsilon$$

for all $N\ge N_0$. Multiplication with $2q_N$ yields

$$2q_N(L-\epsilon)<2p_N<2q_N(L+\epsilon)$$

for all $N\ge N_0$. The logarithm is a strictly monotonically increasing function. Thus previous inequality implies

$$\ln2q_N(L-\epsilon)<\ln 2p_N<\ln 2q_N(L+\epsilon)$$

and

$$\ln2q_N+\ln(L-\epsilon)<\ln 2p_N<\ln 2q_N+\ln (L+\epsilon).$$

$\ln 2q_N>0$ for sufficiently large $N$, thus we may divide the inequality by this quantity to get

$$1+\frac{\ln(L-\epsilon)}{\ln2q_N} <\frac{\ln2p_N}{\ln2q_N}<1+\frac{\ln(L+\epsilon)}{\ln2q_N}$$

for all $N\ge N_0$. We take the limit as $N\to \infty$ and get (9).

To show the opposite direction assume that

$$\frac{p_N}{q_N}$$

is unbounded. Then there exists an $N_0$ such that

$$p_N>q_N$$

for all $n\ge N_0$. Next observe that for every $0<\epsilon<1$ we have $x>x^{1-\epsilon}$ for all $x>1$. This implies that

$$p_N>q_N>q_N^{1-\epsilon}$$

for all $N\ge N_0$. Thus

$$\frac{\ln p_N}{\ln q_N} >\frac{\ln q_N^{1-\epsilon}}{\ln q_N} =(1-\epsilon)$$

for all $N\ge N_0$. Since this inequality holds for all $0<\epsilon<1$ it follows that

$$\frac{\ln p_N}{\ln q_N}\ge 1.$$

This implies the contrapositive of $(\ref{cond1})\Rightarrow (\ref{cond2})$. If $\lim_{N\to\infty}\frac{p_N}{q_N}=0$, we apply the same argument to $\frac{q_N}{p_N}$. This completes the proof of the Corollary.

$\square$

This last Corollary cannot be extended to the case $p=1$. In that case convergence of a rearrangement is equivalent to the condition

$$0<\lim_{N\to\infty}\frac{\ln2 p_N}{\ln2 q_N}<\infty$$ (10)

by Corollary 2. However, (10) is much weaker than (8). To see this consider the case when $p_N=N^2$ and $q_N=N$. Then

$$\lim_{N\to\infty}\frac{\ln2 p_N}{\ln2 q_N}=\lim_{N\to\infty}\frac{\ln2+2\ln N}{\ln2+\ln N}=2,$$

and

$$\frac{p_N}{q_N}=N,$$

which is unbounded. These sequences do not satisfy $p_N+q_N=N$, but the sequences $q_N=[\sqrt{N}]$ and $p_N=N-q_N$ do and allow us to construct a rearrangement of the series

$$\sum_{N=2}^{\infty} \frac{(-1)^n}{n\ln n}$$

which converges to a finite number with unbounded $\frac{p_N}{q_N}$.

We finish this section by studying the series

$$\sum_{n=2}^{\infty}\frac1{\ln n}(-1)^n.$$

A straight forward application of Theorem 1 gives that a rearrangement with a sequence of positive terms $p_N$ and negative terms $q_N$ converges to a finite number, if and only if

$$\lim_{N\to\infty}\int_{2q_N}^{2p_N}\frac1{\ln t}\,dt=\lim_{... ...ty}\left( \mathop{\rm Li}(2p_N)-\mathop{\rm Li}(2q_N)\right)$$

exists and is finite. Here $Li(x)$ denotes the logarithmic integral function defined as

$$\mathop{\rm Li}(x)=\int_2^x\frac1{\ln t}\,dt.$$

Using l'Hospital's rule we get

$$\lim_{x\to\infty}\frac{\mathop{\rm Li}(x)\ln x}{x}=1,$$

and therefore the rearrangement converges if and only if

$$\lim_{N\to\infty}\left(\frac{2p_N}{\ln 2p_N}-\frac{2q_N}{\ln 2q_N}\right)$$

exists and is finite. Factoring this expression as before we get

$$\lim_{N\to\infty}\frac{2q_N}{\ln 2q_N}\left(\frac{p_N\ln 2q_N}{q_N\ln 2p_N}-1\right),$$

which can only be finite if

$$\lim_{N\to\infty}\frac{p_N\ln 2q_N}{q_N\ln 2p_N}=1 .$$ (11)

From the proof of the last Corollary we have that if $\lim_{N\to\infty}\frac{p_N}{q_N}=L$ for some finite positive number $L$, then

$$\lim_{N\to\infty}\frac{\ln 2p_N}{\ln 2q_N}=1 .$$

It follows that in this case (11) holds if and only if $L=1$. However, (11) could hold if $\lim_{N\to\infty}\frac{p_N}{q_N}=0$ or if it diverges. To investigate this case let $r_N=\frac{p_N}{q_N}$, and assume that $r_N\to\infty$ then

$$\frac{p_N\ln 2q_N}{q_N\ln 2p_N}=r_N\frac{\ln r_N2p_N}{\ln 2 p_N}=r_N\left(\frac{\ln r_N}{\ln 2p_N}+1\right) \ge r_N.$$

Clearly, this expression diverges if $r_N\to\infty$. On the other hand if $r_N\to 0$ we investigate the reciprocal of this expression in the same way. Therefore (11) can never be satisfied. We have thus shown

Corollary 5   A rearrangement of

$$\sum_{n=2}^{\infty}\frac1{\ln n}$$

converges if and only if

$$\lim_{N\to\infty}\frac{p_N}{q_N}=1.$$

This last result has a slightly different interpretation in light of the prime number theorem. Let $\pi(x)$ denote the number of primes that are less than or equal to $x$, then the prime-number theorem [1, p. 74] implies that a rearrangement of this series converges if and only if

$$\lim_{N\to\infty}(\pi(2p_N)-\pi(2q_N))$$

is finite. Or in other words if $A_N$ is the number of primes in the interval $[2q_N,2p_n]$, then the rearrangement converges if and only if $\lim_{N\to\infty}A_N$ is finite.