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.

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.
