Assign a value to an infinite divergent series

Question

I am looking for a way to assign a value to an infinite divergent series, but so far I have failed to introduce my series in a more "Riemann-zeta"-lish way.

The series is:

$\sum_{n=0}^{∞}(10^n)$ = 1 + 10 + 100 + 1000 + ...

I am not applying the Term Test, since this series is obviously divergent.

I wouldn't bother myself with this problem, but take the series as infinity, however this nasty little Riemman-zeta function gives me anxiety, about what the actual result of this series is.

As it is well known,

10 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1,

100 = 1 + 1 + 1 + 1 + 1 + ... + 1 + 1 + 1 + 1 + 1

and so on, so we can break down every term of the series like this and "feed" it within the series. But if we feed it that way, the result will be $\zeta(0)=-\frac{1}{2}$ and this I guess is not correct, since the same approach for

$\zeta(-1)=\sum_{n=0}^{∞} \frac{1}{n^{-1}} = \frac{1}{1^{-1}} + \frac{1}{2^{-1}} + \frac{1}{3^{-1}} + ... = 1 + 2 + 3 + 4 + ... = 1 + (1 + 1) + (1 + 1 + 1) + ... = -\frac{1}{12}$

will lead us to the idea, that $\zeta(-1)=\zeta(0)$.

I am not far from the conclusion, that the main algebraic motives like associativity, distributivity and commutativity worth nothing here, as we deal with infinities.

The deeper I go to the problem, the more I am likely to drown in it, when I read about the other Zeta functions, the Cesaro summation, Abel summation, Ramanujan summation, Power series and who knows how many more nearly related to this problem approaches.

Things get even more dim as I read the previously raised similar questions

Is it possible to assign a value to the sum of primes?

Why does $1+2+3+\cdots = -\frac{1}{12}$?

and it seems to me, that there is too much unsolved Philosophy within the Math community, about what happens, when we obtain a value, that sort of equals "out of nowhere" to a divergent series.

I am like pretty sure this series has some weird unsolicited, non-integer, small negative value, but who knows what this value is like?

Can somebody help me out in here?

$\sum_{n=0}^{∞}(10^n)$ = ?

Thanks in advance!

Answer 1

This is an often-confusing bit of abuse of language.

Something like Cesaro "summation" isn't really (when we dive into the formalism) a way of taking the sum of a divergent series. Rather, it should be thought of as a partial function $\mathfrak{C}$ from the set $\mathbb{R}^\mathbb{N}$ of all infinite sequences of real numbers to $\mathbb{R}$ satisfying certain properties, among them the following:

Whenver $(a_n)_{n\in\mathbb{N}}$ is a sequence such that $\sum_{n\in\mathbb{N}}a_n$ is defined (in the usual sense) we have $\mathfrak{C}((a_n)_{n\in\mathbb{N}})=\sum_{n\in\mathbb{N}}a_n$.

That is, our "summation method" extends classical summation. It shouldn't be thought of as actually taking the sum of a divergent series, but rather assigning a number to it in a certain way which we hope is interesting and valuable (and in particular is compatible with the usual notion of summation).

We use sloppy terminology because $(i)$ it's more convenient and $(ii)$ with a little experience no confusion results, but what's actually going on isn't mysterious at all; there's no philosophical import here.

In particular, it is not the case in general that $$\zeta(z)=\sum_{n\in\mathbb{N}}{1\over n^z};$$ rather, the $\zeta$ function is a particular analytic function which agrees with $\sum_{n\in\mathbb{N}}{1\over n^z}$ when the latter is defined (that is, when $Re(z)>1$) but "does its own thing" elsewhere.

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To get a sense of how the more precise language above works, consider the following:

There is no $\mathfrak{D}:\mathbb{R}^\mathbb{N}\rightarrow\mathbb{R}$ which is defined on all of $\mathbb{R}^\mathbb{N}$, extends the usual notion of summation, and satisfies the associativity laws $$\mathfrak{D}((a_n)_{n\in\mathbb{N}})=\mathfrak{D}(a_0+a_1, a_2+a_3, a_4+a_5, ...)$$ and $$\mathfrak{D}((a_n)_{n\in\mathbb{N}})=\mathfrak{D}(a_0, a_1+a_2, a_3+a_4, ...).$$

HINT: consider the Grandi series.

Answer 2

Very easy.

$$\sum_{n=0}^{\infty}10^n=\frac{9\ 10^{\delta (0)-\frac{1}{2}}}{\log ^2(10)}-\frac{1}{9}$$

If you are interested in the regularized value, it is

$$\operatorname{reg}\sum_{n=0}^{\infty}10^n=\frac{1}{\log (10)}-\frac{1}{9}$$