Find the value of the series: $$1-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}+\frac{1}{3}-\frac{1}{10}-\frac{1}{12}-\frac{1}{14}-\frac{1}{16}+\frac{1}{5}\cdots$$
I know that the alternated harmonic series $$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}$$ converges to $\ln2$, but here the order of the terms is different.
On
Let us elaborate on @Arthur idea by regrouping the terms five by five.
$$S=\sum_{k=0}^\infty \left(\frac{1}{2k+1}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6}-\frac{1}{8k+8}\right)$$
Rewriting $\frac{1}{2k+1}=\frac{4}{4(2k+1)}=\frac{4}{8k+4}$ simplifies the sum to four terms
$$S=\sum_{k=0}^\infty \left(-\frac{1}{8k+2}+\frac{3}{8k+4}-\frac{1}{8k+6}-\frac{1}{8k+8}\right)$$
that have all even denominator, so
$$S=-\frac{1}{2}\sum_{k=0}^\infty \left(\frac{1}{4k+1}-\frac{3}{4k+2}+\frac{1}{4k+3}+\frac{1}{4k+4}\right)$$
The relation
$$\int_0^1 x^ndx=\frac{1}{n+1} $$
changes the fractions into integrals $$S=-\frac{1}{2}\sum_{k=0}^\infty \int_0^1 \left(x^{4k}-3x^{4k+1}+x^{4k+2}+x^{4k+3}\right)dx$$
Now we may interchange the order of summation and integration and use the sum of the geometric progression
$$\sum_{k=0}^\infty cx^{ak+b} = \frac{cx^b}{1-x^a}$$
to obtain
$$\begin{align} -2S&=\int_0^1 \sum_{k=0}^\infty \left(x^{4k}-3x^{4k+1}+x^{4k+2}+x^{4k+3}\right)dx\\ &=\int_0^1 \frac {1-3x+x^2+x^3}{1-x^4} dx \\ &=\int_0^1 \frac {1-2x-x^2}{(1+x)(1+x^2)} dx \\ &=\int_0^1 \left( \frac{1}{1+x}-\frac{2x}{1+x^2}\right)dx\\ &=\left(\log(2)-\log(2)\right)=0\\ \end{align}$$
The procedure here is very similar to one of the four proofs Jack D'Aurizio gives to this question, which includes the one here.
This is a special case of a result by Fon Brown, L. O. Cannon, Joe Elich, and David G. Wright, On Rearrangements of the Alternating Harmonic Series, The College Mathematics Journal, Vol. 16, No. 2. (Mar., 1985), pp. 135-138.