Since the alternating harmonic series
$$ \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \frac11-\frac12+\frac13-\frac14+\cdots
$$
is convergent but not absolutely convergent, any real number can be obtained by suitable re-arrangement and grouping of the terms. But finding a re-arrangement that yields a specific real number can be a challenge.
- Find a grouping/re-arrangement that sums to 1.
- Find a grouping/re-arrangement that sums to 0.
And my real question is:
- Find a grouping/re-arrangement that sums to $\sqrt{2}$.
Later edit:
I believe the re-arrangement leading to any particular real value $\alpha$ is not unique. For example, split the original series into two parts, one with terms $\frac11, -\frac12, \frac15, -\frac16, \frac19, -\frac1{10} \cdots$ and the other with the remaining terms. Each of these can be re-arranged to form $\frac12 \log 2$. By inserting those re-arrangements in the order first term from first group, first term from second group, second term from first group,
and so forth, you get a re-arrangement of terms adding to $\log 2$, which is distinct from the "obvious" trivial re-arrangement.
Best Answer
There are many rearrangements that give you the desired value in each case. But when does there exist a rearrangement with a consise description?
Rearrangement for zero. $$ \sum _{k=1}^{\infty }\big({\frac { -1}{8\,k-6}}+{ \frac { -1}{8\,k-4}}+{\frac { -1}{8\,k-2}}+{\frac { -1 }{8 k}}+{\frac { +1}{2\,k-1}}\big)=0 $$ Four evens followed by one odd, repeat.
added
Proof by asymptotics. As $K \to \infty$,
$$ s_0(K):=\sum_{k=1}^K \frac{1}{8k} = \frac{\log K}{8} + \frac{\gamma}{8}+o(1) \\ s_2(K):=\sum_{k=1}^K \frac{1}{8k-2} = \frac{\log K}{8} + \frac{\gamma}{8}+\frac{3\log 2}{8} - \frac{\pi}{16}+o(1) \\ s_4(K):=\sum_{k=1}^K\frac{1}{8k-4} = \frac{\log K}{8} + \frac{\gamma}{8} +\frac{\log 2}{4} + o(1) \\ s_6(K) :=\sum_{k=1}^K \frac{1}{8k-6} = \frac{\log K}{8} + \frac{\gamma}{8}+\frac{3\log 2}{8} + \frac{\pi}{16}+o(1) \\ t_1(K):=\sum_{k=1}^K \frac{1}{2k-1} = \frac{\log K}{2} + \frac{\gamma}{2}+\log 2 + o(1) $$ Then add $$ -s_0(K)-s_2(K)-s_4(K)-s_6(K)+t_1(K) = o(1) $$