THE MAIN RESULT

To state and prove the main result we will first introduce some notation, namely let

$$\{a_k\}_{k=1}^{\infty}$$

be a non-negative sequence, that converges to zero and is eventually decreasing.Moreover, let $f$ be a continuous, non-negative, and eventually decreasing function on $[1,\infty)$ such that

$$f(k) =a_k,$$

for all positive integers $k$. Such a function will always exist, since one can just take the piecewise linear function connecting the points $(k,a_k)$. Define

$$F(x)=\int_1^xf(t)\,dt.$$

The alternating series

$$T=\lim_{N\to\infty}T_N=\lim_{N\to\infty}\sum_{k=1}^Na_k(-1)^{k-1}$$

converges by the alternating series test. If in addition $F$ is bounded the series converges absolutely.

To continue let

$$S_N(A)$$ (2)

be the $N$-th partial sum of a rearrangement of the series that converges to a real number $A$, and let $\{p_N\}_{N=1}^{\infty}$ and $\{q_N\}_{N=1}^{\infty}$ be the related sequences of positive and negative terms mentioned above. It is easily checked that

$$p_N+q_N=N,$$

and

$$S_N(A)=\sum_{k=1}^{p_N}a_{2k-1}-\sum_{k=1}^{q_N}a_{2k}.$$

We can now state the main result of this note:

Theorem 1   With the notations introduced above we have:

$$\lim_{N\to\infty}\left(F(2p_N)-F(2q_N)\right)= 2A-2T$$ (3)

Proof: Without loss of generality, we will assume that $p_N\ge q_N$ for all sufficiently large values of $N$. The other case can be shown in the completely analogous way. We have

$$\begin{eqnarray*} S_N(A)&=&\sum_{k=1}^{p_N}a_{2k-1}-\sum_{k=1}^{q_N}a_{2k}\\ ... ...+1}^{p_N}a_{2k-1}\\ &=&T_{2q_N}+\sum_{k=q_N+1}^{p_N}a_{2k-1}, \end{eqnarray*}$$

where $T_{2q_N}$ is the $2q_N$-th partial sum for the original alternating series. From Figure 1 below we see that

$$2\sum_{k=q_N+2}^{p_N}a_{2k-1}\le \int_{2q_N}^{2p_N}f(t)\,dt,$$

and therefore

$$2\sum_{k=q_N+1}^{p_N}a_{2k-1}-2a_{2q_N+1}\le \int_{2q_N}^{2p_N}f(t)\,dt.$$ (4)

Figure 2 shows that

$$2\sum_{k=q_N}^{p_N}a_{2k-1}\ge \int_{2q_N}^{2p_N}f(t)\,dt,$$

and therefore

$$2\sum_{k=q_N+1}^{p_N}a_{2k-1}+2a_{2q_N-1}\le \int_{2q_N}^{2p_N}f(t)\,dt.$$ (5)

Figure 1: Upper estimate of the sum by the integral

        

Figure 2: Lower estimate of the sum by the integral

Combining (4) and (5), we obtain

$$\int_{2q_N}^{2p_N}f(t)\,dt-2a_{2q_N-1} -2a_{2q_N-1}\le 2\su... ...N+1}^{p_N}a_{2k-1}\le \int_{2q_N}^{2p_N}f(t)\,dt+2a_{2q_N+1},$$

which immediately implies

$$\int_{2q_N}^{2p_N}f(t)\,dt-2a_{2q_N-1} -2a_{2q_N-1}\le 2S_N(A)-2T_{2q_N}\le \int_{2q_N}^{2p_N}f(t)\,dt+2a_{2q_N+1}.$$ (6)

Finally, we observe that

$$\lim_{N\to\infty}2a_{2q_N-1}=\lim_{N\to\infty}2a_{2q_N+1}=0.$$

Therefore the desired result follows from taking the limit as $N\to \infty$ in all terms of (6).

$\square$

Corollary 1   If $\lim_{x\to\infty} F(x)$ exists and is finite, the series is absolutely convergent, and

$$\lim_{N\to\infty} \left(F(2p_N)-F(2q_N)\right)=0.$$

Therefore, any rearrangement of an absolutely convergent series converges to the same limit.

Proof: In this case we have

$$\lim_{N\to\infty}F(2p_N)=\lim_{N\to\infty}F(2q_N),$$

and the result follows immediately.

$\square$