Are we allowed to rearrange a convergent series by grouping elements of it and converging to the same number. for a sequence like $\sum_{0}^{\infty}a_{n} = a_{0} + a_{1} + ... = L$
are we allowed to group its element like $b_{0} = a_{0} + a_{1} = a_{1} + a_{0}$  and  $b_{1} = a_{2} + a_{3} = a_{3} + a_{2}$ , and so on. (in pairs)
then can we conclude that;
$\sum_{0}^{\infty}a_{n} = \sum_{0}^{\infty}b_{n} = \sum_{0}^{\infty}a_{n} = (a_{1} + a_{0}) + (a_{3} + a_{2})... = L$
I know that even if I can not have such a rearrangement, still $\sum_{0}^{\infty}a_{n} = L \Rightarrow   (a_{1} + a_{0}) + (a_{3} + a_{2})... = L $ holds. So my question is not that do they converge to the same real number L. My question is that are we allowed to make such an rearrangement and say they converge to the same real number L? And if this rearrangement is doable then I wonder whether I can generalize it by grouping, say 7 many elements, by their permutations?
 A: If a series $\sum a_n$ converges to $L$, then the series $\sum b_n$ with $b_n = a_{kn}+ a_{kn +1}+ \dots + a_{kn+k-1}$ also converges to $L$, whatever the integer $k>1$ is.
This is not difficult to prove as $\{a_n\}$ converges to zero.
Obviously, the converse is not true.
A: No, we cannot. Because the infinite series are a bit tricky to handle. Unlike finite summations, the infinite summations can lead to contradictions if rearranged. Please, refer to this link for a detailed answer. Check out Ramanujan Summation of Natural Numbers. While proving that the $1-1+1-1+\cdots = \frac{1}{2}$. He has rearranged the summation which lead to the contradiction(because the limit of the sum is indeterminate). One more example. If we assume that the above statement is true then: 
\begin{align}
S = 1+\left({1 \over 2}   + {1 \over 3}\right)+\left({1 \over 4}   + {1 \over 5}\right)\cdots   \\= 1+ {1 \over 3}   + \left({1 \over 2} +{1 \over 4}\right)   + {1 \over 5} \cdots \\
= 1+ \left({1 \over 3}   \right) +\left[\left({1 \over 2} + {1 \over 4}\right)   + {1 \over 5}\right] \cdots
\end{align}
We can do this till we don't get
\begin{align}
S = 1+{1 \over 3}   + {1 \over 5}+{1 \over 7}+ \cdots + \left({1 \over 2}   + {1 \over 4}+{1 \over 6} \cdots \right ) \\
= 1+{1 \over 3}   + {1 \over 5}+{1 \over 7}+ \cdots + {1 \over 2}\left({1 \over 1}   + {1 \over 2}+{1 \over 3} \cdots \right )
\end{align}
$$ S =  \left(1+{1 \over 3}   + {1 \over 5}+{1 \over 7}+ \cdots \right)+{1 \over 2} S $$
From the above we conclude $ {1 \over 2} S  ={1 \over 2}   + {1 \over 4}+{1 \over 6} \cdots = 1+{1 \over 3}   + {1 \over 5}+{1 \over 7}+ \cdots $ . Which doesn't seems to be obvious.
Also, See the  Link here
