I have a question about summation methods. A value is assigned to a divergent sum. All methods agree that $ \texttt{super-}\sum_{k=1}^{\infty} k^p = -\frac{B_{p+1}^{+}}{p+1}$ where $ B_{p}$ are Bernoulli numbers.

My questions are the following:

- Can we define a ‘superlimit’ to assign values to divergent sequences? Can we do it in such a way that a supersum is the superlimit of its partial sums. Below is how to do this for polynomials.
- Is there a reason why there are many methods to do supersums and seamingly none that do superlimits. Is there something inherrently wrong with the concept of a superlimit that is not wrong with supersums?

A suggestion for a superlimit for polynomials is: $ $ \texttt{superlimit}_{k\to\infty} k^p = \int_{0}^{1} k^p \texttt{dn} \Bigl( = \frac{1}{p+1}\Bigr)$ $

Now $ $ \int_{0}^{1} \Bigl( \sum_{k=1}^{n-1} k^p \Bigr) \texttt{dn} = \int_{0}^{1} \frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_{j}^{-}n^{p-j+1} \texttt{dn} $ $ $ $ = \frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_{j}^{-}\frac{1}{p-j+2} $ $ $ $ = -\frac{B_{p+1}^{+}}{p+1}$ $ through Faulhabers formula. Which proves that this choice of superlimit implies supersums that agree with the zeta function. This proof was in the answer to this post here.

The converse can be proven inductively. So if we want the superlimit of a polynomial agree with the known supersum values $ \texttt{super-}\sum_{k=1}^{\infty} k^p = -\frac{B_{p+1}^{+}}{p+1}$ and cut off the partial sums at $ N-1$ then $ $ \texttt{superlimit}_{k\to\infty} k^{p} = \frac{1}{p+1}$ $

If the answer to the first questions is ‘yes we can define a superlimit’ then I also have the following question:

- When assigning values to divergent sequences should we not also give alternative values to sequences that go to zero?

It is clear that $ $ \texttt{superlimit}_{n\to\infty}(a_n \cdot b_n) \neq \texttt{superlimit}_{n\to\infty}(a_n) * \texttt{superlimit}_{n\to\infty}(b_n)$ $ (where $ \cdot$ is the inner product of the sequences).

But I think it should hold that $ $ \texttt{superlimit}(1/a_n) = 1/{\texttt{superlimit}(a_n)}$ $

If this is the case then $ \texttt{superlimit}(1/k^{p}) = p+1$ if $ p>=0$