$-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac{1}{10}...$
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?
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!
As has been noted, rearranging a conditionally convergent series needs to be done with care. Collecting terms into contiguous groups whose size is uniformly bounded is allowed, as long as the terms tend to $0$.
Assuming that the series can be continued as
$$
\begin{align}
&\left(-\frac12-\frac14\right)+\left(\frac13-\frac16-\frac18\right)+\left(\frac15-\frac1{10}-\frac1{12}\right)+\cdots\tag{1a}\\
&=-\frac34+\sum_{k=1}^\infty\left(\frac1{2k+1}-\frac1{4k+2}-\frac1{4k+4}\right)\tag{1b}\\
&=-\frac34+\sum_{k=1}^\infty\left(\frac1{4k+2}-\frac1{4k+4}\right)\tag{1c}\\
&=-\frac34+\frac12\sum_{k=1}^\infty\left(\frac1{2k+1}-\frac1{2k+2}\right)\tag{1d}\\
&=-1+\frac12\sum_{k=0}^\infty\left(\frac1{2k+1}-\frac1{2k+2}\right)\tag{1e}\\
&=-1+\frac12\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\tag{1f}\\
&=-1+\frac12\log(2)\tag{1g}
\end{align}
$$
Explanation:
$\text{(1a):}$ collect the terms into contiguous groups whose size is no greater than $3$
$\text{(1b):}$ write the sum using sigma notation
$\text{(1c):}$ simplify the terms
$\text{(1d):}$ pull a factor of $\frac12$ outside the sum $\left(\sum ca_k=c\sum a_k\right)$
$\text{(1e):}$ add the $k=0$ term to the sum and subtract $\frac14$ from $-\frac34$ to compensate
$\text{(1f):}$ recognize the sum as an alternating series (from uniformly sized groups)
$\text{(1g):}$ apply the Alternating Harmonic Series
The Alternating Harmonic Series
$$
\begin{align}
\sum_{k=0}^\infty\frac{(-1)^k}{k+1}
&=\lim_{n\to\infty}\sum_{k=0}^n\frac{(-1)^k}{k+1}\tag{2a}\\
&=\lim_{n\to\infty}\sum_{k=0}^n\int_0^1(-1)^kx^k\,\mathrm{d}x\tag{2b}\\
&=\lim_{n\to\infty}\int_0^1\sum_{k=0}^n(-1)^kx^k\,\mathrm{d}x\tag{2c}\\
&=\lim_{n\to\infty}\int_0^1\frac{1+(-1)^nx^{n+1}}{1+x}\,\mathrm{d}x\tag{2d}\\
&=\int_0^1\frac1{1+x}\,\mathrm{d}x+\lim_{n\to\infty}(-1)^n\int_0^1\frac{x^{n+1}}{1+x}\,\mathrm{d}x\tag{2e}\\
&=\int_0^1\frac1{1+x}\,\mathrm{dx}\tag{2f}\\[6pt]
&=\log(2)\tag{2g}
\end{align}
$$
Explanation:
$\text{(2a):}$ definition of an infinite sum
$\text{(2b):}$ $\int_0^1(-1)^kx^k\,\mathrm{d}x=\frac{(-1)^k}{k+1}$
$\text{(2c):}$ a finite sum of integrals is the integral of the finite sum
$\text{(2d):}$ evaluate the finite geometric sum
$\text{(2e):}$ an integral of a finite sum is the finite sum of the integrals
$\phantom{\text{(2e):}}$ a limit of a finite sum is the finite sum of the limits
$\text{(2f):}$ $0\le\int_0^1\frac{x^{n+1}}{1+x}\,\mathrm{d}x\le\int_0^1x^{n+1}\,\mathrm{d}x=\frac1{n+2}$
$\phantom{\text{(2f):}}$ thus, $\lim\limits_{n\to\infty}(-1)^n\int_0^1\frac{x^{n+1}}{1+x}\,\mathrm{d}x=0$
$\text{(2g):}$ evaluate the integral
You cannot rearrange an infinite series which is not absolutely convergent. We can evaluate it as follows. Your infinite series is \begin{align} \sum_{n=1}^{\infty}\left(-\dfrac1{4n-2} - \dfrac1{4n}+\dfrac1{2n+1}\right) & = \sum_{n=1}^{\infty}\left(-\dfrac1{4n-2} - \dfrac1{4n}+ 2\cdot \dfrac1{4n+2}\right)\\ & = \sum_{n=1}^{\infty}\left(-\int_0^1x^{4n-3}dx- \int_0^1x^{4n-1}dx + 2\int_0^1x^{4n+1}dx\right)\\ & = \sum_{n=1}^{\infty}\int_0^1 x^{4n}\left(2x-\dfrac1x - \dfrac1{x^3}\right)dx\\ & = \int_0^1\left(2x-\dfrac1x - \dfrac1{x^3}\right)\left(\sum_{n=1}^{\infty}x^{4n}\right)dx\\ & = \int_0^1\left(2x-\dfrac1x - \dfrac1{x^3}\right) \dfrac{x^4}{1-x^4}dx\\ & = \int_0^1 \left(\dfrac{x}{x^2+1}-2x\right)dx = \dfrac{\ln(2)}2-1 \end{align}
To justify the swapping of the limit and integral, we have \begin{align} I_m & = \sum_{n=1}^m \int_0^1 x^{4n}\left(2x-\dfrac1x-\dfrac1{x^3}\right)dx = \int_0^1 \dfrac{x^4\left(1-x^{4m}\right)}{1-x^4}\left(2x-\dfrac1x-\dfrac1{x^3}\right)dx\\ & = \int_0^1 \left(\dfrac{x}{x^2+1}-2x\right)dx + \int_0^1 \dfrac{x(2x^2+1)}{x^2+1}\cdot x^{4m}dx \end{align}
Now $$\lim_{m\to\infty} \int_0^1 \dfrac{x(2x^2+1)}{x^2+1}\cdot x^{4m}dx = 0$$ There are multiple ways to prove this. One is to just invoke dominated convergence theorem, since $x^{4m} \leq 1$ and $\displaystyle \int_0^1 \dfrac{x(2x^2+1)}{x^2+1}dx < \infty$. Other is to simply note that $\dfrac{x(2x^2+1)}{x^2+1} \leq \dfrac32$ on $[0,1]$. Hence, $$\lim_{m\to\infty} \int_0^1 \dfrac{x(2x^2+1)}{x^2+1}\cdot x^{4m}dx \leq \lim_{m\to\infty} \int_0^1 \dfrac32\cdot x^{4m}dx = \lim_{m\to\infty} \dfrac3{2(4m+1)} = 0$$
Hence, $$\lim_{m \to \infty} I_m = \dfrac{\log(2)}2-1$$
