showing $ 1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0 $ How to show that the following infinite series
$$
1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0?
$$
The above series is of the form $\sum_{n \ge 1} \frac{f(n)}{n}$, where $f$ is a periodic arithmetical function of period $4$, with the values $f(1)=f(3)=f(4)=1$ and $f(2)=-3$. Since $\sum_{1 \le i \le 4} f(i)=0$, it is assured that this series is convergent.
 A: I'm a little wary of doing this, since it's not absolutely convergent, but can you rewrite each period as
$$
\left(\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{4k-1}-\frac{1}{4k}\right) -
\left(\frac{2}{4k-2}-\frac{2}{4k}\right)
$$
which can in turn be rewritten as
$$
\left(\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{4k-1}-\frac{1}{4k}\right) -
\left(\frac{1}{2k-1}-\frac{1}{2k}\right)
$$
and then you have two parallel alternating harmonic series, both of which sum to $\ln 2$.  Subtract one from the other and you get $0$.
But, as I say, I'm not sure that's kosher.
A: Your series is
$$\begin{align}
S &=\sum_{k=0}^\infty \left( \frac{1}{4k+1}-\frac{3}{4k+2} + \frac{1}{4k+3} + \frac{1}{4k+4}\right)\\
&=\sum_{k=0}^\infty \int_0^1 (x^{4k}-3x^{4k+1}+x^{4k+2}+x^{4k+3})dx\\
&=\int_0^1 \left( \sum_{k=0}^\infty  \left(x^{4k}-3x^{4k+1}+x^{4k+2}+x^{4k+3}\right) \right)dx\\
&=\int_0^1 \frac{1-3x+x^2+x^3}{1-x^4} dx\\
&=\int_0^1 \left(\frac{1}{1+x}-\frac{2x}{1+x^2}\right)dx\\
&=\log(2)-\log(2)\\
&=0\\
\end{align}
$$
and this zero relation was the motivation for OEIS sequence http://oeis.org/A176563
It can also be constructed from $2\log(2)-\log(4)$ as follows
$$\begin{align}
2\log(2)
&=2\sum_{k=0}^\infty \left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)\\
&=2\sum_{k=0}^\infty \left(\frac{1}{4k+1}-\frac{1}{4k+2}+\frac{1}{4k+3}-\frac{1}{4k+4}\right)\\
&=\sum_{k=0}^\infty \left(\frac{2}{4k+1}-\frac{2}{4k+2}+\frac{2}{4k+3}-\frac{2}
{4k+4}\right)\\
\log(4)&=\sum_{k=0}^\infty \left(\frac{1}{4k+1}+\frac{1}{4k+2}+\frac{1}{4k+3}-\frac{3}{4k+4}\right)\\ 
2\log(2)-\log(4)&=\sum_{k=0}^\infty \left(\frac{2-1}{4k+1}-\frac{2+1}{4k+2}+\frac{2-1}{4k+3}-\frac{2-3}{4k+4}\right)\\
&=\sum_{k=0}^\infty \left(\frac{1}{4k+1}-\frac{3}{4k+2}+\frac{1}{4k+3}+\frac{1}{4k+4}\right)\\
\end{align}$$
For the formula for $\log(4)$, see Do these series converge to logarithms?
A: sum of all odd terms: $X = (1 + 1/3 + 1/5 + ... )$
sum of all terms (4n+2): $Y = -3/2(1 + 1/3 + 1/5 + ...) = -3X/2$
Sum of remaining terms = $Z = 1/4 + 1/8 + 1/12 + ... = 1/4(1 + 1/2 + 1/3 + 1/4 + 1/5 + ...)$
splitting into odd and even terms: $Z = 1/4(1+ 1/3 + 1/5 +...) + 1/8(1 + 1/2 + 1/3 + ...)$
So $Z = X/4 + Z/2 \implies Z = X/2$
So $X + Y + Z = X - 3X/2 + X/2 = 0$
