Rearrange the series $ \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$ to converge to $1$. I have studied the Riemann's theorem about rearrangement of conditionally convergent series. Also I have seen other rearrangements of the given series on this site that converge to different sums $\ln2,\;\frac{3}{2}\ln2 $, etc. But Iam not able to visualize the rearrangement that converges to $1$. Please help. Thank You.
 A: The positive terms are $$ 1, \frac 1 3, \frac 1 5, \frac 1 7, \frac 1 9, \frac 1 {11}, \frac 1 {13}, \ldots $$ Their sum diverges to $+\infty$.  The negative terms are $-1$ multiplied by $$ \frac 1 2, \frac 1 4, \frac 1 6, \frac 1 8, \frac 1 {10}, \frac 1 {12}, \ldots $$ Their sum diverges to $-\infty$.
$1 + \dfrac 1 3$ exceeds $1$.  Now add enough negative terms to that to get a sum less than $1$:
$$
1 + \frac 1 3 - \frac 1 2 = \frac 5 6 < 1.
$$
Then add enough positive terms after that to make the sum more than $1$:
$$
1 + \frac 1 3 - \frac 1 2 + \frac 1 5 = \frac{31}{30} >1.
$$
Then add enough negative terms after that to make the sum less than $1$:
$$
1 + \frac 13 - \frac12 + \frac 1 5 - \frac 1 4 = \frac{47}{60} <1.
$$
Then add enough positive terms after that to make the sum more than $1$:
$$
1 + \frac 13 - \frac12 + \frac 1 5 - \frac 1 4 \underbrace{{} + \frac 1 7 + \frac 1 9} = \frac{1307}{1260} > 1.
$$
This last time we needed two terms.  Is there a pattern to the number of terms we have to add at each step? Maybe not.  But we know that it will always be possible to make the sum more than $1$ or less than $1$ as the case may be, because the series of positive terms and the series of negative terms both diverge to infinity.
A: We can do it like this.  Following each even term, add either one or two odd terms.  For $M=0,1,2,3,\dots$ we want at stage $M$ a total of $M$ even terms and $\lfloor M e^2/4 \rfloor$ odd terms.  Since $1 < e^2/4 < 2$, when $M$ increases by $1$, the value $\lfloor M e^2/4 \rfloor$ increases by either $1$ or $2$.
This arrangement of the series has sum $1$.
Explanation.
For odd terms, as $N \to \infty$ we have partial sums
$$
A_N := \sum_{n=0}^N \frac{1}{2n+1} = \log(2\sqrt{N}\,) + \frac{\gamma}{2}+O(N^{-1})\qquad\text{as }N \to \infty
\tag1$$
($\gamma$ is the Euler-Mascheroni constant.)
For even terms, as $M \to \infty$ we have partial sums
$$
B_M := \sum_{n=1}^M\frac{-1}{2n} = -\log(\sqrt{M}\,) - \frac{\gamma}{2} + O(N^{-1})
\tag2$$
Now, using the scheme described above, at stage $M$, the partial sum of our rearranged series is
$B_M + A_N$ where $N = \lfloor M e^2/4 \rfloor$.  So
$$
B_M+A_N = \log\frac{2\sqrt{N}}{\sqrt{M}} + O(M^{-1})
\tag3$$
Note that
$$
\frac{Me^2}{4}-1 < N \le \frac{Me^2}{4}
\\
e^2 - \frac{4}{M} < \frac{4 N}{M} \le e^2
$$
As $M \to \infty$,
$$
\frac{4N}{M} \to e^2
\\
\frac{2\sqrt{N}}{\sqrt{M}} \to e
\\
\log\frac{2\sqrt{N}}{\sqrt{M}} \to 1
\\
B_M + A_N \to 1 .
$$
