A while ago, there was a great hype about the “identity”
$$\sum_{n=1}^{\infty} n = -\frac{1}{12}.$$
Apart from some series manipulations where the validity seems to be at least questionable, the derivation of this always goes through the zeta function:
Where the series converges, the zeta function is defined by
$$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s}$$
and outside that range by analytic continuation. And it turns out that inserting $s=-1$ formally results in
$$\zeta(-1) = -\frac{1}{12} = \sum_{n=1}^\infty n$$
However looking at the series in isolation, there is no indication that the zeta function should be chosen.
An obvious way to get an analytic function that at one point gives the sum of all natural numbers is
$$f(x) = \sum_{n=1}^{\infty} nx^n$$
at $x=1$, however (not surprisingly) that function diverges at $1$.
Therefore my question:
Is it possible to get another finite value for the series by analytic continuation of another series?
Concretely, do there exist continuous functions $f_1, f_2, f_3, \ldots$ such that
On some non-empty open subset $S$ of $\mathbb C$, $f(x)=\sum_{n=1}^\infty f_n(x)$ converges to an analytic function.
At some point $x_0$, $f_n(x_0) = n$ for all positive integers $n$.
The analytic continuation of $f$ is well defined and finite at $x_0$.
$f(x_0) \ne -1/12$
What if we demand the functions $f_n$ to be analytic rather than just continuous?
Answer
What if we demand the functions $f_n$ to be analytic rather than just continuous?
No problem. Define
$$f_n(s) = \frac{n}{(n - (-1)^n)^s},$$
where $k^s$ is as usual defined using the real value of $\log k$ (works since $n - (-1)^n > 0$). Then $f_n(0) = n$ for all $n$, and by a standard argument the series converges absolutely and locally uniformly for $\operatorname{Re} s > 2$. We compute
\begin{align}
\sum_{n = 1}^{\infty} f_n(s)
&= \sum_{n = 1}^{\infty} \frac{n}{(n - (-1)^n)^s} \\
&= \sum_{n = 1}^{\infty} \Bigl(\frac{1}{(n - (-1)^n)^{s-1}} + \frac{(-1)^n}{(n - (-1)^n)^s}\Bigr) \\
&= \sum_{n = 1}^{\infty} \frac{1}{(n - (-1)^n)^{s-1}} + \sum_{n = 1}^{\infty} \frac{(-1)^n}{(n - (-1)^n)^s} \\
&= \biggl(\frac{1}{2^{s-1}} + \frac{1}{1^{s-1}} + \frac{1}{4^{s-1}} + \frac{1}{3^{s-1}} + \ldots\biggr) \\
&\qquad + \biggl(-\frac{1}{2^s} + \frac{1}{1^s} - \frac{1}{4^s} + \frac{1}{3^s} - \frac{1}{6^s} + \frac{1}{5^s} - \ldots\biggr) \\
&= \sum_{m = 1}^{\infty} \frac{1}{m^{s-1}} + \sum_{m = 1}^{\infty} \frac{(-1)^{m-1}}{m^s} \\
&= \zeta(s-1) + \eta(s)
\end{align}
for $\operatorname{Re} s > 2$. This has an analytic continuation to $\mathbb{C}\setminus \{2\}$, and the value at $0$ is
$$\zeta(-1) + \eta(0) = -\frac{1}{12} + \frac{1}{2} = \frac{5}{12}.$$
One can by similar means obtain different values.
Such summation methods are however very ad-hoc, as far as I know every "reasonable" summation method assigns either $+\infty$ (the natural value) or $-\frac{1}{12}$ to the divergent series. I admit that I don't know a good definition of "reasonable" for summation methods (except maybe "extends 'limit of partial sums', is linear and stable", but that definition excludes several widely used summation methods).
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