# Flipping the order of an infinite series and assigning it a value?

I was recently wondering if there was some method to "flip" the series and still assign it a value for divergent series. To illustrate what I mean:

$$S(x,n) = a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n$$

We would like to define the flipped sum series:

$$F(S(x,n)) = a_n + a_{n-1} x^{1} + \dots + a_0 x^{n}$$

Hence, to do this we begin by: assigning $x \to x^{-1}$ and take $x^n$ common:

$$S(x^{-1},n) = (a_n + a_{n-1} x^{1} + \dots + a_0 x^{n}) x^{-n}$$

Multiplying both sides with $x^n$:

$$S(x^{-1},n) x^n = (a_n + a_{n-1} x^{1} + \dots + a_0 x^{n})$$

Taking limit $n \to \infty$:

$$\lim_{n \to \infty}S(x^{-1},n) x^n = \lim_{n \to \infty}(a_n + a_{n-1} x^{1} + \dots + a_0 x^{n})$$

Taking the second limit $x \to 1$:

$$\lim_{x \to 1 } \lim_{n \to \infty}S(x^{-1},n) x^n = \lim_{n \to \infty} (a_n + a_{n-1} + \dots + a_0 )$$

## Questions

Is the above proof correct? Can this somehow be extended to zeta regularization?

• What happened to the limit with respect to $n$ on the right hand side of your last equation? – Bungo Sep 20 '17 at 16:16
• My bad. It was a typo ... Thanks to your comment It also made me realize I may have to swap the order of the limits. But I'm not sure ... – drewdles Sep 20 '17 at 16:23
• In your example, is $a_0 + a_1 + ...$ divergent? If that's the case, then "flipping" around doesn't change the partial sums in any way, so the divergent behaviour won't go away. – theindigamer Sep 20 '17 at 16:32
• Indeed the swapping of the limits needs justification as it's not valid in general. I agree with @theindigamer that even if the result holds, it doesn't seem to help, assuming $a_0 + a_1 + \cdots$ is divergent. Your final expression is exactly the same as $\lim_{n \to \infty}(a_0 + a_1 + \cdots + a_n)$ since addition of finitely many numbers is commutative. – Bungo Sep 20 '17 at 16:47
• Your question is a nonsense. Do you understand analytic continuation ? Flipping a finite sum is trivial, flipping a series doesn't make sense. You need to look at the axioms of a summation method $T(\sum_{n \ge 1} a_n) = C$. Then the power series regularization is defined by $T(\sum_{n \ge 1} a_{n+1}) = T(\sum_{n \ge 1} a_n)-a_1$ (shift invariance) and $T(\sum_{n \ge 1} a_{n})=T(\sum_{n \ge 1} a_{n/m})$. It is not the case for the zeta regularization which is defined by $T(\sum_{n \ge 1} a_{n})=T(\sum_{n \ge 1} a_{n^{1/m}})=T(\sum_{n \ge 1} a_{n/m}) = C$ (multiplicative invariance) – reuns Sep 21 '17 at 10:45