Two equivalent series converge on different limits Consider the series:
$$
\sum_{n=0}^{\infty} \left({\dfrac{(-1)^n}{2n+1}}\right)^3  = \dfrac{1}{1^3}-\dfrac{1}{3^3}+\dfrac{1}{5^3}-\dfrac{1}{7^3} \dots = \dfrac{\pi^3}{32}
$$
Now suppose I wish to represent this series with the following equivalent summation:
$$
\sum_{n=1}^{\infty} \left[\left({\dfrac{1}{4n-3}}\right)^3-\left({\dfrac{1}{4n-1}}\right)^3 \right]=  \left[\dfrac{1}{1^3}-\dfrac{1}{3^3}\right]+\left[\dfrac{1}{5^3}-\dfrac{1}{7^3}\right] \dots 
$$
I know that changing order of summation in infinite series might affect the limit, but in this example the order of summation is not really changed, and all of the terms are exactly the same.One can even prove by induction that both series term by term are equivalent - i.e that the second is a compressed version of the first.
So can I state that the limit of the second series is equal to $\dfrac{\pi^3}{32}$ ? If not, I would be happy to find some explanation as to why we even pretend to assign values to infinite series if we can see that certain examples of infinite series break basic axioms of arithmetic.
 A: What seems to be in question here (as commented by Mourad) is the Riemann Rearrangement Theorem, which says that if a series converges, but does not converge absolutely, then the terms of that series can be rearranged to converge to any real number.
This series converges absolutely; that is, the series of its absolute values converges:
$$
\begin{align}
\sum_{n=0}^\infty\left|\left(\frac{-1}{2n+1}\right)^3\right|
&=\sum_{n=0}^\infty\frac1{(2n+1)^3}\\[3pt]
&=\frac78\zeta(3)
\end{align}
$$
If a series converges absolutely, its terms can be reordered (and regrouped) in any way desired and the series will converge to the same limit.

As mentioned by José Carlos Santos, if a series converges, even conditionally,  the partial sums of any grouping of the terms will simply give a subsequence of the partial sums of the original series. Thus, any grouping of the terms will give the same limit of partial sums.
However, if the original series does not converge, then regrouping terms might produce different results. For example,
$$
\sum_{n=0}^\infty(-1)^n
$$
does not converge, and its terms can be grouped as
$$
(1-1)+(1-1)+(1-1)+\dots=0
$$
or as
$$
1-(1-1)-(1-1)-(1-1)-\dots=1
$$
A: If a series$$a_0+a_1+a_2+a_3+a_4+a_5+\cdots$$converges and its sum is $S$, then the series$$(a_0+a_1)+(a_2+a_3)+(a_4+a_5)+\cdots$$converges too and its sum is also $S$. That's so because the sequence of partial sums of the second series is a subsequence of the sequence of partial sums of the original one. So, the answer is affirmative.
