How to calculate the series $-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac{1}{10}...$?

$-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac{1}{10}...$

After rearrangement the series looks like $\sum^{\infty}_{n=2}\frac{(-1)^{n+1}}{n}$.

My way of doing this is using Taylor series of $\ln{(1+x)}=\sum^{\infty}_{n=1}\frac{(-1)^{n+1}x^n}{n}$.

Therefore let $x=1$, $\sum^{\infty}_{n=2}\frac{(-1)^{n+1}}{n}=\sum^{\infty}_{n=1}\frac{(-1)^{n+1}x^n}{n}-\sum^{1}_{n=1}\frac{(-1)^{n+1}x^n}{n}=\ln{2}-\sum^{1}_{n=1}\frac{(-1)^{n+1}x^n}{n}=\ln{2}-1$

Is my solution correct?

Solution 1:

Firstly, your method is wrong as has been pointed out. You cannot rearrange an infinite series, because the definition of its value is the limit of its partial sums, and rearranging it anyhow will change the sequence of partial sums, so there is no guarantee that the limit will be the same.

Just for a concrete simple illustration:

$\frac12 - \frac13 + \frac14 - \frac15 \cdots$

$\ = ( \frac12 - \frac13 ) + ( \frac14 - \frac15 ) + \cdots$   [Note that even putting brackets here is not trivially true!]

$\ > 0$.

$- \frac13 - \frac15 + \frac12 - \frac17 - \frac19 + \frac14 - \frac1{11} - \frac1{13} + \frac16 \cdots$

$\ = ( - \frac13 - \frac15 + \frac12 ) + ( - \frac17 - \frac19 + \frac14 ) + ( - \frac1{11} - \frac1{13} + \frac16 ) + \cdots$

$\ = ( - \frac{3+5}{3 \cdot 5} + \frac12 ) + ( - \frac{7+9}{7 \cdot 9} + \frac14 ) + ( - \frac{11+13}{11 \cdot 13} + \frac16 ) + \cdots$

$\ < ( - \frac{3+5}{4 \cdot 4} + \frac12 ) + ( - \frac{7+9}{8 \cdot 8} + \frac14 ) + ( - \frac{11+13}{12 \cdot 12} + \frac16 ) + \cdots$

$\ = 0$.

Both series converge, so clearly rearrangement may not even preserve the sign of the sum.

Secondly, we can find the limit of your series quite easily.

We just need the following (loose) inequality for any $m,n \in \mathbb{N}_{>1}$ such that $m \le n$:

$\ln(n+1) - \ln(m) = \int_m^{n+1} \frac1x\ dx < \sum_{k=m}^n \frac1k < \int_{m-1}^n \frac1x\ dx = \ln(n) - \ln(m-1)$.

We first simplify every third partial sum:

$-\frac12-\frac14+\frac13 -\frac16-\frac18+\frac15 -\cdots -\frac1{4n+2}-\frac1{4n+4}+\frac1{2n+3}$

$\ = - ( \frac12+\frac14 + \cdots + \frac1{4n+2}+\frac1{4n+4} ) + ( \frac13+\frac15 + \cdots + \frac1{2n+3} )$

$\ = - ( \frac12+\frac14 + \cdots +\frac1{4n+4} ) + \left( ( \frac11+\frac12 + \cdots + \frac1{2n+3} ) - ( \frac12+\frac14 + \cdots + \frac1{2n+2} ) - 1 \right)$

$\ = -\frac12 ( \frac11+\frac12 + \cdots + \frac1{2n+2} ) + \left( ( \frac11+\frac12 + \cdots + \frac1{2n+3} ) - \frac12 ( \frac11+\frac12 + \cdots + \frac1{n+1} ) \right) - 1$

$\ = \frac12 ( \frac11+\frac12 + \cdots + \frac1{2n+2} ) - \frac12 ( \frac11+\frac12 + \cdots + \frac1{n+1} ) - 1 + \frac1{2n+3}$

$\ = \frac12 \sum_{k=n+2}^{2n+2} \frac1k - 1 + \frac1{2n+3}$

$\ \in \frac12 [ \ln(2n+3)-\ln(n+2) , \ln(2n+2)-\ln(n+1) ] - 1 + \frac1{2n+3}$

$\ = \frac12 [ \ln(2-\frac{1}{n+2}) , \ln(2) ] - 1 + \frac1{2n+3}$.

Note that we only rearranged a finite sum, not an infinite series.

Now clearly as $n \to \infty$:

$-\frac12-\frac14+\frac13 -\frac16-\frac18+\frac15 -\cdots -\frac1{4n+2}-\frac1{4n+4}+\frac1{2n+3}$

$\ \to \frac12 \ln(2) - 1$   [because $\frac1{2n+3} \to 0$ and $\frac{1}{n+2} \to 0$ and $\ln$ is continuous at $2$].

Since the terms of the original series tend to $0$, it also converges to $\frac12 \ln(2) - 1$. Done!