The other night I was considering the way in which we can split a finite sum of any arithmetic function into two finite sums, one for it's odd and another for even index terms :
$$\sum _{k=1}^{n} \left( -1 \right) ^{k}a_{{k}}=\sum _{k=1}^{ \lfloor \frac{n}{2} \rfloor }a_{{2\,k}}-\sum _{k=0}^{\lfloor \frac{n}{2} \rfloor -\delta \left( \frac{n}{2},\lfloor \frac{n}{2} \rfloor \right) +1}a_{{2\,k+1}} $$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(FS001)}$$
where:
$\delta \left( x,y \right) =\cases{1&$x=y$\cr 0&$x\neq y$\cr}$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(KD001)}$$
And then I decided to be a little brave ask myself if this equality would hold for an infinite number of natural numbers:
$$\lim _{n\rightarrow \infty}\Bigl(\sum _{k=1}^{n} \left( -1 \right) ^{k}a_{{k}}\Bigr)=\lim _{n\rightarrow \infty}\Bigl(\sum _{k=1}^{ \lfloor \frac{n}{2} \rfloor }a_{{2\,k}}\Bigr)-\lim _{n\rightarrow \infty}\Bigl(\sum _{k=0}^{\lfloor \frac{n}{2} \rfloor -\delta \left( \frac{n}{2},\lfloor \frac{n}{2} \rfloor \right) +1}a_{{2\,k+1}} \Bigr)$$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(IL001)}$$
Therefore the equality for the indefinite sum as follows:
$$\sum _{k=1}^{\infty} \left( -1 \right) ^{k}a_{{k}}=\sum _{k=1}^{\infty}a_{{2\,k}}-\sum _{k=0}^{\infty}a_{{2\,k+1}} $$
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(IL002)}$$
A preliminary investigation has lead to me holding a belief that this will be true in the event that $\sum_{k=1}^{\infty}a_{{k}}$ is a convergent sum, (therefore, as too are the sums on the right hand side of our equality above) however I have this unsettling feeling that even if for other arithmetic functions I use finite sum approximations to their divergent sums, and these results do indeed appear to confirm that the equality holds in the infinite limit, this will not always be the case and there will exist pathalogical residuals that are dependant on the specific arithmetic function in which they occur.
So my questions are:
1)Does this bad feeling I have about the indefinite extension of the equality have any merit to warrant further study?
2) If these non zero residual functions exist, would they serve as a good foundation for me to construct predicates a partition of the set of all arithmetic functions based on the nature of their from this equality?
Be nice please I don't like this subject very much.
If $a_k=(-1)^{k}\frac 1 k$ then all the three series under consideration are divergent so the equation is not valid. Howoever, if $\sum a_k$ and $\sum (-1)^{k}a_k$ are both assumed to be convergent then everything is fine.