I am studying the series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$, which converges to $\ln 2$. By rearranging the terms (one positive term followed by two negative terms), we obtain a new series:
$$1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \frac{1}{12} + \frac{1}{7} - \frac{1}{14} + \cdots$$
Grouping terms, this simplifies to: $$\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \cdots = \frac{\ln 2}{2}.$$
However, the convergence of the rearranged series is not immediately clear. A more rigorous argument is provided as follows:
Let $s_n$ be the sum of the first $n$ terms of the rearranged series. Taking three terms at a time, we have: $$s_{3n} = \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \cdots + \frac{1}{4n-2} - \frac{1}{4n} = \frac{1}{2} \sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k}.$$
Thus, $s_{3n} \rightarrow \frac{\ln 2}{2}$. Additionally, the $n$-th term of the rearranged series tends to zero as $n \rightarrow \infty$. The text concludes that $s_n \rightarrow \frac{\ln 2}{2}$ by "an easy argument left to the reader."
My Question:
How can we rigorously prove that $s_n \rightarrow \frac{\ln 2}{2}$? Specifically, how does the fact that the $n$-th term tends to zero, combined with the behavior of $s_{3n}$, allow us to conclude that the entire sequence $s_n$ converges to $\frac{\ln 2}{2}$?
some effort:
Behavior of $s_{3n+1}$ and $s_{3n+2}$
The sequence $s_n$ is not just $s_{3n}$; it also includes $s_{3n+1}$ and $s_{3n+2}$. To show that $s_n \rightarrow \frac{\ln 2}{2}$, we need to analyze these cases as well.
- Case 1: $s_{3n+1}$
The term $s_{3n+1}$ adds one more term to $s_{3n}$. Specifically:
$s_{3n+1} = s_{3n} + \frac{1}{4n+1}.$
Since $\frac{1}{4n+1} \rightarrow 0$ as $n \rightarrow \infty$, we have:
$ s_{3n+1} = s_{3n} + \frac{1}{4n+1} \rightarrow \frac{\ln 2}{2} + 0 = \frac{\ln 2}{2}. $
- Case 2: $s_{3n+2}$
The term $s_{3n+2}$ adds two more terms to $s_{3n}$. Specifically:
$ s_{3n+2} = s_{3n} + \frac{1}{4n+1} - \frac{1}{4n+2}. $
Since both $\frac{1}{4n+1} \rightarrow 0$ and $\frac{1}{4n+2} \rightarrow 0$ as $n \rightarrow \infty$, we have:
$ s_{3n+2} = s_{3n} + \frac{1}{4n+1} - \frac{1}{4n+2} \rightarrow \frac{\ln 2}{2} + 0 - 0 = \frac{\ln 2}{2}. $ but again I cant understand where is the "easy argument" Maybe I forgot some theorem?
