This question is related to an earlier post.

Suppose we have a sequence $(a_n)$ of real numbers and we are interested in the existence and value of the series \begin{equation} \lim_{N \to \infty} \qquad \sum^N_{n = 1} (a_n - a_{n-1}) \\,. \end{equation} Looking at the partial sum, we have $S_N = a_N - a_0$.

Now here comes my question:

Is it always (i.e. for any sequence $(a_n)$) true that \begin{equation} \lim_{N \to \infty} \qquad \sum^N_{n = 1} (a_n - a_{n-1}) = \lim_{N \to \infty} a_N - a_0 \qquad ? \end{equation}

I am asking because in the book "Yet another Introduction to Analysis" by Victor Bryant there is the following remark, suggesting the answer to my above question is NO (not always):

If we let $S_N = \sum_{n = 1}^N (-1)^{n+1} \frac{1}{n}$ then the resulting series exists. Denote its value by $S$.

Writing the same partial sum as \begin{equation} T_N = 1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \dots \text{up to } (-1)^{N+1}\frac{1}{N} \end{equation} (i.e we rearrange the terms) we have $T_{3N} = \frac{1}{2} S_{2N}$, hence

\begin{equation} T = \lim_{N \to \infty} T_N = \lim_{N \to \infty} T_{3N} = \frac{1}{2}\lim_{N \to \infty} S_{2N} = \frac{1}{2} \lim_{N \to \infty} S_N = \frac{1}{2}S \,. \end{equation}

So all we did was rearranging the summands in the partial sums before taking the limit, and it changed the value of the series.

To repeat my question:

Is it always (i.e. for any sequence $(a_n)$) true that \begin{equation} \lim_{N \to \infty} \qquad \sum^N_{n = 1} (a_n - a_{n-1}) = \lim_{N \to \infty} a_N - a_0 \qquad ? \end{equation}

Many thanks for your help!


I thought I answered this at the other question. $T_N$ is not a rearrangement of $S_N$. Pick a value of $N$, and write out $S_N$, and write out $T_N$, and you will see that there are terms in the one that are not in the other.

  • $\begingroup$ Ah! Now I understand, this is where my confusion comes from, because the example actually talks about different sequences of partial sums. I am sorry I didn't get your answer in the first place! $\endgroup$ – harlekin Jan 22 '12 at 12:13

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