I will be considering the Riemann Zeta Function to assign values to some limits.
Values of $\zeta(s)$ at negative integer arguments are known.
For example,
$$\zeta(0)=-\frac{1}{2}\stackrel{\Re}{=}\sum_{n=1}^{\infty}1$$ $$\zeta(-1)=-\frac{1}{12}\stackrel{\Re}{=}\sum_{n=1}^{\infty}n$$
This got me wondering what would happen if we take the Partial Sums first and then put the limit as follows:
(I will be omitting the $\Re$ above the equality sign but assume we are only assigning values here)
$$-\frac{1}{2}=\lim_{n\to\infty}n$$
$$-\frac{1}{12}=\lim_{n\to\infty}\frac{n(n+1)}{2}$$
Now we can eliminate $n$ from the Second Limit and find the value for $n^2$.
$$\lim_{n\to\infty}n^2=\frac{1}{3}$$
I tried this process further and got the following examples,
$$\lim_{n\to\infty}n^3=-\frac{1}{4}$$
$$\lim_{n\to\infty}n^4=\frac{1}{5}$$
So it seems that,
$$\lim_{n\to\infty}n^{s}=\frac{(-1)^s}{1+s}$$
Can we prove this regularization?
I was able to solve it by reducing it to Bernoulli Numbers:
It is known that,
$$\sum_{k=1}^nk^c=\frac{1}{1+c}\sum_{k=0}^c\binom{c+1}{k}B_kn^{c+1-k}$$
So if we take the limit as $n\to \infty$ and assign the values.
$$\zeta(-c)=\frac{1}{1+c}\sum_{k=0}^c\binom{c+1}{k}\frac{B_k(-1)^{c+1-k}}{1+c+1-k}$$
But,
$$\zeta(-c)=-\frac{B_{1+c}}{1+c}$$
This gives us.
$$-B_{n}=(-1)^n\sum_{k=0}^{n-1}(-1)^{k}\binom{n}{k}\frac{B_k}{n+1-k}$$
Which is a recurrence for Bernoulli Numbers, mentioned on Wikipedia.