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.
[Math] Rearrange the series $ \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$ to converge to $1$.
convergence-divergencesequences-and-series
Best Answer
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.