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?

 A: 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
A: It is a rewarding  exercise to write, or read,  detailed rigorous proofs of  some simple, "obvious" results. 
The definition of $x=\sum_{j=1}^{\infty}y_j$ is that $x=\lim_{n\to \infty}S_n$ where $S_n=\sum_{j=1}^ny_j.$   A useful  way to state (or define) that the sequence $S=(S_n)_n$ converges to $x$  is that for any $r>0 $ the set $$F(S, r)=\{n: S_n\not \in [-r+x,r+x]\}$$ is a finite set.
Let  $x=\sum_{j=1}^{\infty}y_j.$  Insert some $0$'s into the sequence $(y_j)_j$ to produce a new sequence $(z_i)_i.$ For each $i$ we have either $z_i=0$ or $z_i=y_{g(i)}$ where $g(i)\leq i.$
Let $T=(T_m)_m$ where $T_m=\sum_{i=1}^mz_i.$
Let $i_0$ be the least (or any) $i$ such that $z_i=y_1.$ 
Then for every $m\geq i_0$ we have $T_m=\sum_{  \{g(i):i\leq m\}}y_{g(i)}= S_{f(m)}$  where $f(m)=\max \{g(i):i\leq m\}.$
Note that for any $n$ there exists $m_0$ such that $z_{m_0}=y_{n+1}$ so there exists $m_0$ such that $f(m_0)=n+1.$
Now for any $r>0$ we have  $$F(T,r)=\{m:T_m\not \in [-r+x,r+x]\}\subset \{m\geq i_0:S_{f(m)} \not \in [-r+x.r+x]\}\cup \{m:m<i_0\}$$ $$=\{m\geq i_0: f(m)\in F(S,r)\}\cup \{m:m<i_0\}.$$  Observe that for any $n$ the set $\{m\geq i_0: f(m)=n\}$ is a finite set, because there exists $m_0$ such that $f(m_0)=n+1,$ and hence $m\geq m_0\implies f(m)>n.$ And recall that $F(S,r)$ is a  finite set because the sequence $S$ converges to $x.$
So for any $r>0$ the set $F(T,r)$ is a subset of a union of a finite collection of finite sets: $$F(T,r)\subset (\cup_{n\in F(S,r)}\{m\geq  i_0:f(m)=n\})\cup \{m:m<i_0\}.$$ Therefore $F(T,r)$ is finite for every $r>0$. Therefore $T$ converges to $x.$ Therefore $x=\sum_{i=1}^{\infty}z_i.$
A: *

*Yes, that is ok.

*Yes it is. The partial sum of $\sum_k (a_k+b_k)$ is the partial sum of $\sum_ka_k$ plus the partial sum of $\sum_k b_k.$ The result follows from the sum property for limits.

*Yes. Adding zeros will only delay the inevitable convergence of the sequence of partial sums. Where you insert zeros, the sequence of partial sums will hold flat. For the $N$ you find in the proof of convergence of the original, simply replace with $N$ plus the number of zeros you inserted before the N-th and you'll have the same value that will be within $\epsilon$ of the number the sum converges to.

