Euler summation

In the mathematics of convergent and divergent series, Euler summation is a summability method. That is, it is a method for assigning a value to a series, different from the conventional method of taking limits of partial sums. Given a series ∑an, if its Euler transform converges to a sum, then that sum is called the Euler sum of the original series. As well as being used to define values for divergent series, Euler summation can be used to speed the convergence of series.

Euler summation can be generalized into a family of methods denoted (E, q), where q ≥ 0. The (E, 1) sum is the ordinary Euler sum. All of these methods are strictly weaker than Borel summation; for q > 0 they are incomparable with Abel summation.

Definition

For some value y we may define the Euler sum (if it converges for that value of y) corresponding to a particular formal summation as:

${\displaystyle _{E_{y}}\,\sum _{j=0}^{\infty }a_{j}:=\sum _{i=0}^{\infty }{\frac {1}{(1+y)^{i+1}}}\sum _{j=0}^{i}{\binom {i}{j}}y^{j+1}a_{j}.}$

If the formal sum actually converges, an Euler sum will equal it. But Euler summation is particularly used to accelerate the convergence of alternating series and sometimes it can give a useful meaning to divergent sums.

To justify the approach notice that for interchanged sum, Euler's summation reduces to the initial series, because

${\displaystyle y^{j+1}\sum _{i=j}^{\infty }{\binom {i}{j}}{\frac {1}{(1+y)^{i+1}}}=1.}$

This method itself cannot be improved by iterated application, as

${\displaystyle _{E_{y_{1}}}{}_{E_{y_{2}}}\sum =\,_{E_{\frac {y_{1}y_{2}}{1+y_{1}+y_{2}}}}\sum .}$

Examples

• Using y = 1 for the formal sum
${\displaystyle \sum _{j=0}^{\infty }(-1)^{j}P_{k}(j)}$
we get
${\displaystyle \sum _{i=0}^{k}{\frac {1}{2^{i+1}}}\sum _{j=0}^{i}{\binom {i}{j}}(-1)^{j}P_{k}(j),}$
if Pk is a polynomial of degree k. Note that the inner sum would be zero for i > k, so in this case Euler summation reduces an infinite series to a finite sum.
• The particular choice
${\displaystyle P_{k}(j):=(j+1)^{k}}$
provides an explicit representation of the Bernoulli numbers, since
${\displaystyle {\frac {B_{k+1}}{k+1}}=-\zeta (-k)}$
(the Riemann zeta function). Indeed, the formal sum in this case diverges since k is positive, but applying Euler summation to the zeta function (or rather, to the related Dirichlet eta function) yields
${\displaystyle {\frac {1}{1-2^{k+1}}}\sum _{i=0}^{k}{\frac {1}{2^{i+1}}}\sum _{j=0}^{i}{\binom {i}{j}}(-1)^{j}(j+1)^{k}}$
which is of closed form.
• ${\displaystyle \sum _{j=0}^{\infty }z^{j}=\sum _{i=0}^{\infty }{\frac {1}{(1+y)^{i+1}}}\sum _{j=0}^{i}{\binom {i}{j}}y^{j+1}z^{j}={\frac {y}{1+y}}\sum _{i=0}^{\infty }\left({\frac {1+yz}{1+y}}\right)^{i}}$
With an appropriate choice of y (i.e. equal to or close to −1/z) this series converges to 1/1 − z.