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?
 A: 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
A: 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,
$$\lim_{m \to \infty} I_m = \dfrac{\log(2)}2-1$$
