Ramanujan summation

Since there are no properties of an entire sum, the Ramanujan summation functions as a property of partial sums. If we take the Euler–Maclaurin summation formula together with the correction rule using Bernoulli numbers, we see that:

${\displaystyle {\begin{aligned}{\frac {1}{2}}f(0)+f(1)+\cdots +f(n-1)+{\frac {1}{2}}f(n)&={\frac {f(0)+f(n)}{2}}+\sum _{k=1}^{n-1}f(k)=\sum _{k=0}^{n}f(k)-{\frac {f(0)+f(n)}{2}}\\&=\int _{0}^{n}f(x)\,dx+\sum _{k=1}^{p}{\frac {B_{2k}}{(2k)!}}\left[f^{(2k-1)}(n)-f^{(2k-1)}(0)\right]+R_{p}\end{aligned}}}$

Ramanujan[1] wrote it for the case p going to infinity, and changing the limits of the integral and the corresponding summation:

${\displaystyle \sum _{k=a}^{x}f(k)=C+\int _{a}^{x}f(t)\,dt+{\frac {1}{2}}f(x)+\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(x)}$

where C is a constant specific to the series and its analytic continuation and the limits on the integral were not specified by Ramanujan, but presumably they were as given above. Comparing both formulae and assuming that R tends to 0 as x tends to infinity, we see that, in a general case, for functions f(x) with no divergence at x = 0:

${\displaystyle C(a)=\int _{0}^{a}f(t)\,dt-{\frac {1}{2}}f(0)-\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(0)}$

where Ramanujan assumed ${\displaystyle a=0.}$ By taking ${\displaystyle a=\infty }$ we normally recover the usual summation for convergent series. For functions f(x) with no divergence at x = 1, we obtain:

${\displaystyle C(a)=\int _{1}^{a}f(t)\,dt+{\frac {1}{2}}f(1)-\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(1)}$

C(0) was then proposed to use as the sum of the divergent sequence. It is like a bridge between summation and integration.

The most common application of Ramanujan summation is for the Riemann zeta function ζ(z), in which the Ramanujan summation of the function ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{n^{s}}}}$ has the same value as ζ(s) for all the values of ${\displaystyle s}$, even for those for which the first function is divergent, which is equivalent to doing analytic continuation or, alternatively, applying smoothed sums.[2]

The convergent version of summation for functions with appropriate growth condition is then:

${\displaystyle f(1)+f(2)+f(3)+\cdots =-{\frac {f(0)}{2}}+i\int _{0}^{\infty }{\frac {f(it)-f(-it)}{e^{2\pi t}-1}}\,dt}$

In the following text, ${\displaystyle ({\mathfrak {R}})}$ indicates "Ramanujan summation". This formula originally appeared in one of Ramanujan's notebooks, without any notation to indicate that it exemplified a novel method of summation.

For example, the ${\displaystyle ({\mathfrak {R}})}$ of 1 − 1 + 1 − ⋯ is:

${\displaystyle 1-1+1-\cdots ={\frac {1}{2}}\quad ({\mathfrak {R}}).}$

Ramanujan had calculated "sums" of known divergent series. It is important to mention that the Ramanujan sums are not the sums of the series in the usual sense,[3][4] i.e. the partial sums do not converge to this value, which is denoted by the symbol ${\displaystyle ({\mathfrak {R}}).}$ In particular, the ${\displaystyle ({\mathfrak {R}})}$ sum of 1 + 2 + 3 + 4 + ⋯ was calculated as:

${\displaystyle 1+2+3+\cdots =-{\frac {1}{12}}\quad ({\mathfrak {R}})}$

Extending to positive even powers, this gave:

${\displaystyle 1+2^{2k}+3^{2k}+\cdots =0\quad ({\mathfrak {R}})}$

and for odd powers the approach suggested a relation with the Bernoulli numbers:

${\displaystyle 1+2^{2k-1}+3^{2k-1}+\cdots =-{\frac {B_{2k}}{2k}}\quad ({\mathfrak {R}})}$

It has been proposed to use of C(1) rather than C(0) as the result of Ramanujan's summation, since then it can be assured that one series ${\displaystyle \textstyle \sum _{k=1}^{\infty }f(k)}$ admits one and only one Ramanujan's summation, defined as the value in 1 of the only solution of the difference equation ${\displaystyle R(x)-R(x+1)=f(x)}$ that verifies the condition ${\displaystyle \textstyle \int _{1}^{2}R(t)\,dt=0}$.[5]

This demonstration of Ramanujan's summation (denoted as ${\displaystyle \textstyle \sum _{n\geq 1}^{\mathfrak {R}}f(n)}$) does not coincide with the earlier defined Ramanujan's summation, C(0), nor with the summation of convergent series, but it has interesting properties, such as: If R(x) tends to a finite limit when x → 1, then the series ${\displaystyle \textstyle \sum _{n\geq 1}^{\mathfrak {R}}f(n)}$ is convergent, and we have

${\displaystyle \sum _{n\geq 1}^{\mathfrak {R}}f(n)=\lim _{N\to \infty }\left[\sum _{n=1}^{N}f(n)-\int _{1}^{N}f(t)\,dt\right]}$

In particular we have:

${\displaystyle \sum _{n\geq 1}^{\mathfrak {R}}{\frac {1}{n}}=\gamma }$

where γ is the Euler–Mascheroni constant.

Ramanujan resummation can be extended to integrals; for example, using the Euler–Maclaurin summation formula, one can write

${\displaystyle {\begin{aligned}\int _{a}^{\infty }x^{m-s}\,dx&={\frac {m-s}{2}}\int _{a}^{\infty }x^{m-1-s}\,dx+\zeta (s-m)-\sum _{i=1}^{a}\left[i^{m-s}+a^{m-s}\right]\\&\qquad -\sum _{r=1}^{\infty }{\frac {B_{2r}\theta (m-s+1)}{(2r)!\Gamma (m-2r+2-s)}}(m-2r+1-s)\int _{a}^{\infty }x^{m-2r-s}\,dx\end{aligned}}}$

which is the natural extension to integrals of the Zeta regularization algorithm.

This recurrence equation is finite, since for ${\displaystyle m-2r<-1}$,

${\displaystyle \int _{a}^{\infty }dx\,x^{m-2r}=-{\frac {a^{m-2r+1}}{m-2r+1}}.}$

Note that this involves (see zeta function regularization)

${\displaystyle I(n,\Lambda )=\int _{0}^{\Lambda }dx\,x^{n}}$.

With ${\displaystyle \Lambda \to \infty }$, the application of this Ramanujan resummation lends to finite results in the renormalization of quantum field theories.

• Borel summation
• Cesàro summation
• Divergent series
• Ramanujan's sum
• Abel–Plana formula