Arranging the alternating harmonic series to sum to $\sqrt{2}$ 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.
 A: 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
$$
-s_0(K)-s_2(K)-s_4(K)-s_6(K)+t_1(K) = o(1)
$$
A: I don't really think "take the rearrangement given by the standard proof that for every $x$ there exists a rearrangement that sums to $x$" really counts as an explicit rearrangement.
Here's an explicit rearrangement that sums to $0$.
First imagine all the terms arranged in a rectangular grid. The first column consists of all the terms $1/n$ for odd $n$, in decreasing order; each row row looks like $1/n$, $-1/2n$, $-1/4n$, ... Like so:
1   -1/2  -1/4  -1/8 ...
1/3 -1/6  -1/12 -1/24...
1/5 -1/10 -1/20 -1/40 ...
1/7 -1/14 -1/28 -1/56 ...
.
.
. 
Note that each row sums to zero, and we got each term exactly once.
Now traverse that grid in one of the two standard ways that one uses to show that $\Bbb N^2$ is countable. Take the traversal that runs along the boundaries of subsquares. That's a traversal that has the following property:
P: The first $N^2$ terms consist precisely of the first $N$ entries in the first $N$ rows of the grid.
In fact any traversal with property P will work; we can certainly find an "explicit" such traversal if we didn't follow my description of the traversal I have in mind.
Since each row sums to zero and $\sum_{n=N+1}^\infty 2^{-n}=2^{-N}$, the sum of the first $N^2$ terms in the rearranged series is $2^{-N}$ times the sum of $1/n$ over the first $N$ odd values of $n$; this is on the order of $2^{-N}\log(N)$, which tends to zero.
So if $s_n$ is the $n$-th partial sum of the rearrangement the $s_n$ have a subsequence that tends to zero: $$s_{N^2}\to0.$$
Now to show that the series $\sum a_n$ itself sums to $0$ it's enough to show this: $$\lim_{N\to\infty}\sum_{n=N^2+1}^{(N+1)^2}|a_n|=0.$$
That's the sum of the absolute values of the entries in the grid with (i) $1\le\text{ row }\le   N+1,\text{col }=N+1$ and the terms with (ii) $\text{row }=N+1,1\le\text{ col }\le N+1.$
The sum of the terms in (i) is again of the order of $2^{-N}\log N$, while the sum of the terms in (ii) is of the order of $1/N$.
A: Here is how to make a rearrangement that sums up to $1000$, which is a completely arbitrary number, and you may substitute it for any number you like. First of all, let $h_n=\frac{(-1)^{n+1}}n$ be the alternating harmonic sequence in the original order, and let $a_n$ be the rearranged terms in such a way that
$$
a_n=\cases{h_{2m+1} & the first odd term not taken yet if $A_{n-1}\leq 1000$\\h_{2m} & the first even term not taken yet if $A_{n-1} > 1000$}
$$
where $A_{n-1}=\sum_{i=1}^{n-1}a_i$ is the sum of all previous terms.
A: Rearrangement for $\sqrt{2}$  
Let
$$
N_k = \left\lfloor  \frac{e^{2\sqrt{2}}}{4}\;k\right\rfloor ;
\qquad
D_k = N_k - N_{k-1} .
$$
So $(D_k)_{k=1}^{20}$ is
$$
4, 4, 4, 4, 5, 4, 4, 4, 5, 4, 4, 4, 4, 5, 4, 4, 4, 5, 4, 4, 4, 5, 4, 4, 4
$$
and in general, $D_k$ is either $4$ or $5$ in some non-periodic pattern (since $e^{2\sqrt{2}}$ is irrational).  
Define a rearrangement of the series in groups.  Group $k$ consists of one even term and $D_k$ odd terms:
$$
\left(\frac{-1}{2}+\frac{+1}{1}+\frac{+1}{3}+\frac{+1}{5}+\frac{+1}{7}\right)
+
\left(\frac{-1}{4}+\frac{+1}{9}+\frac{+1}{11}+\frac{+1}{13}+\frac{+1}{15}\right)
+\dots
$$
This rearrangement of the series sums to $\sqrt{2}$.
