$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$ converges to any arbitrary real number?

Question

I've been reading about Riemann series theorem and I think I understand the proof and the idea. But still, I have the feeling that something is wrong with it. For instance, if $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=0$ and $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=1$ doesn't this mean $0=1$?

Answer 1

An explanation, rather than a complete proof, of how the reordered series can converge to any predefined limit.

Consider a conditionally convergent series such as the one in your question. The series of positive terms $\sum p_n$ diverges, as does the series of negative terms $\sum q_m$.

Choose $a$ to be whatever limit you want the reordered series—not the original series—to converge to. For the sake of argument let's use $a>0$.

Since $\sum p_n$ diverges, you can take just enough terms from the start of the sequence $(p_n)$ to add up to $>a$. Next, since $\sum q_m$ also diverges, you can follow these with just enough terms from $(q_n)$ to get the sum back below $a$. Switch back to taking positive terms to get above $a$ again.

Continue like this, alternating between groups of positive and negative terms. Each time, the amount by which you overshoot $a$ is no more than the size of the last term used.

Now, since the original series is convergent, the sequences $(p_n)$ and $(q_m)$ both converge to $0$. So the size of the "overshoot" also converges to $0$. In other words, the reordered series we're constructing converges to $a$.

The divergerce of $\sum p_n$ and $\sum q_n$ guarantees you never get stuck one side of $a$, and convergence of $(p_n)$ and $(q_m)$ to $0$ guarantees that the reordered series converges.

The crucial point is that the reordered series isn't the same series, and its limit is the limit of the process of adding more terms—not the sum of "all the positive terms and all the negative terms", which is $\infty-\infty$ and hence undefined.

---

Edit: I think a big source of confusion here is the notation $\sum_{n=0}^{\infty}{a_n}$ which looks as though it should mean "Collect all the infinitely many terms $a_n$ together and find their sum". But it's defined not to mean that because the sum doesn't always exist. Instead it's defined as the limit of a sequence of finite sums containing more and more terms. Changing the order of the terms changes the sequence of sums, which can sometimes change the limit.

Answer 2

The Riemann series theorem has nothing to do with the sum of the series $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}n$ which is equal to $-\log2$ (but it's not equal to $0$ or to $1$).

If you apply the theorem to this series, what it tells you is that, for any $\alpha\in\mathbb R$, there is some bijection $b\colon\mathbb{N}\longrightarrow\mathbb N$ such that $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{b(n)}}{b(n)}=\alpha$.