Infinite zeros in infinite series The problem:

Given that
$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \ldots $$
Prove
$$\frac{\pi}{3} = 1 + \frac{1}{5} - \frac{1}{7} - \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \ldots$$

My solution:
We know
$$
\begin{align}
\frac{\pi}{4} & = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} -\frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \ldots \\
\\
\frac{\pi}{12} & = \frac{1}{3} - \frac{1}{9} + \frac{1}{15} - \frac{1}{21} + \frac{1}{27} -\frac{1}{33} + \frac{1}{39} - \frac{1}{45} + \ldots\\
\\
& = 0 + \frac{1}{3} + 0 + 0 - \frac{1}{9} + 0 + 0 + \frac{1}{15} + 0 + 0 - \frac{1}{21}
\end{align}
$$
now add them together:
$$
\begin{align}
\frac{\pi}{4} + \frac{\pi}{12} & = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \ldots \\
\\
& + 0 + \frac{1}{3} + 0 + 0 - \frac{1}{9} + 0 + 0 + \frac{1}{15} + \ldots \\
\end{align}
$$
and we will get:
$$
\begin{align}
\frac{\pi}{3} & = 1 + 0 + \frac{1}{5} - \frac{1}{7} + 0 -\frac{1}{11} + \frac{1}{13} + 0 + \ldots \\
& =  1 + \frac{1}{5} - \frac{1}{7} -\frac{1}{11} + \frac{1}{13} + \ldots
\end{align}
$$
My questions:


*

*I inserted/removed infinite zeros into/from the series, is that OK?

*My solution relies on the fact that $\Sigma a_n + \Sigma b_n = \Sigma (a_n + b_n)$ and $k \Sigma a_n = \Sigma k a_n$. Is this always true for convergent infinite series? If so, why is it? (yeah I know this is a stupid question, but since I'm adding infinite terms up, I'd better pay some attention.)

*Bouns question: Can I arbitrarily (arbitrariness isn't infinity, you know) insert/remove zeros into/from a convergent infinite series, without changing its convergence value?


Both the given identities follow from the fact that $\frac{\pi}{\text{something}}$ is related with the integral over $(0,1)$ of a rational function. The first identity is a consequence of 
$$\frac{\pi}{4} = \arctan(1)=\int_{0}^{1}\frac{dx}{1+x^2} = \int_{0}^{1}\left(1-x^2+x^4-x^6+\ldots\right)\,dx $$
and for the second series we may perform the same manipulation in the opposite direction, leading to:
$$\begin{eqnarray*}\sum_{n\geq 0}\left(\frac{1}{12n+1}+\frac{1}{12n+5}-\frac{1}{12n+7}-\frac{1}{12n+11}\right)&=&\int_{0}^{1}(1+x^4-x^6-x^{10})\sum_{n\geq 0}x^{12n}\,dx\\&=&\int_{0}^{1}\frac{1+x^4}{1+x^6}\,dx\\&=&\int_{0}^{1}\left(\frac{1}{1+x^2}+\frac{x^2}{1+x^6}\right)\,dx\\(x\mapsto z^{1/3})\qquad &=&\frac{\pi}{4}+\frac{1}{3}\int_{0}^{1}\frac{dz}{1+z^2}=\color{red}{\frac{\pi}{3}}.\end{eqnarray*} $$
Indeed, we are just multipling the first series by $\frac{4}{3}$ :D
