Examples of divergent series summed by means of the analytic continuation of the corresponding

Question

For my Bachelor's thesis, I am investigating divergent series. This is (yet another) question on this topic.

Apparently, a divergent series $$ S = \sum_{n=1}^{\infty} a_{n} $$ can be summed by means of analytic continuation of the corresponding dirichlet series $$ f(s) = \sum_{n=1}^{\infty} \frac{ a_{n} }{n^{s}} $$ at $s=0$.

Howevever, I can not find any examples of divergent series being summed by means of this method (for example, I can't find anything in Hardy's Divergent Series). Can you please provide me with some examples and/or theory which show how this method can be used to sum divergent series? References are also very much appreciated.

Answer 1

For example, for any Dirichlet character $\chi$, the sums $\sum_{n=1}^\infty \chi(n)$ can be summed by analytic continuation of $\sum_n \chi(n)/n^s$, which has a meromorphic continuation.

Similarly, for many other arithmetical functions (such as coefficients of modular forms) the analogous sum has a meromorphic continuation, so is summable in this sense.

However, not every reasonable sequence of coefficients, even if admitting reasonable-looking arithmetic descriptions, gives a Dirichlet series with a meromorphic continuation. Some have natural boundaries, as was discovered by Estermann c. 1928. An account of some relatively elementary examples is in my course notes, linked-to from the HTML with title "Estermann phenomenon", from my course-notes page at http://www.math.umn.edu/~garrett/m/mfms/

In fairly immediate situations, and in general, the question of whether a Dirichlet series has a meromorphic continuation (beyond a fairly obvious half-plane) is a difficult open question.

Answer 2

Example.

The power series $f(x)=\sum_{n=1}^\infty (2^{m_2(n)+1}-1)x^n$, where $m_2(n)$ is the multiplicity of the factor $2$ of $n$ (the $2$-adic order of $n$), diverges for all $x\neq 0$. As $f(x)-2f(x^2)=x+x^2+x^3+\cdots=x/(1-x)$ for $|x|<1$, its iterated elementary Ramanujan sum is $1/2$ (definition, also here).

The Dirichlet series $F(s)=\sum_{n=1}^\infty (2^{m_2(n)+1}-1)n^{-s}$ converges in the half-plane $\sigma>2$. It takes the form $$ F(s)=\frac{1}{1-2^{1-s}}\zeta(s) $$ (here). Therefore, $F(0)=-\zeta(0)=1/2$.

Edit. Change notation: $f_2(x)=f(x)$, $f_1(x)=x+x^2+x^3+\cdots $ and $f_0(x)=x-x^2+x^3-\cdots $.

We know that $f_0(x)=x/(1+x)$ for $|x|<1$ is Abel summable to $\sum_{n=1}^\infty (-1)^{n+1}=1/2$ at $x=1$ (Grandi series) (here), $f_1(x)-2f_1(x^2)=f_0(x)$ and $f_2(x)-2f_2(x^2)=f_1(x)$.

In general, for $f(x)=\sum_{n=0}^\infty a_n x^n$, $f^{(N)}(x)$ is defined by $$ f^{(N)}(x)=\left(x\frac{d}{dx}\right)^N f(x)~. $$

We have $f_1^{(N)}(x)-2^{N+1}f_1^{(N)}(x^2)=f_0^{(N)}(x)$ and $f_2^{(N)}(x)-2^{N+1}f_2^{(N)}(x^2)=f_1^{(N)}(x)$.

We know that the elementary Ramanujan sum of $f_1^{(N)}(x)$ at $x=1$ is $$ f_1^{(N)}(1)=\frac{f_0^{(N)}(1)}{1-2^{N+1}}=\zeta(-N)~. $$

The iterated elementary Ramanujan sum of $f_2^{(N)}(x)$ at $x=1$ is $$ f_2^{(N)}(1)=\frac{f_1^{(N)}(1)}{1-2^{N+1}}=\frac{1}{1-2^{1+N}}\zeta(-N) =F(-N) $$ (also here).