# 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 Σan, 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, 0) sum is the usual (convergent) sum, while (E, 1) 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

Euler summation is particularly used to accelerate the convergence of alternating series and allows evaluating divergent sums.

$_{E_y}\, \sum_{j=0}^\infty a_j := \sum_{i=0}^\infty \frac{1}{(1+y)^{i+1}} \sum_{j=0}^i {i \choose j} y^{j+1} a_j .$

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

$y^{j+1}\sum_{i=j}^\infty {i \choose j} \frac{1}{(1+y)^{i+1}}=1.$

This method itself cannot be improved by iterated application, as

$_{E_{y_1}} {}_{E_{y_2}}\sum = \, _{E_{\frac{y_1 y_2}{1+y_1+y_2}}} \sum.$

## Examples

• We have $\sum_{j=0}^\infty x^j P_k(j)= \sum_{i=0}^k \frac{x^i}{(1-x)^{i+1}}\sum_{j=0}^i {i \choose j} (-1)^{i-j} P_k(j)$, if $P_k$ is a polynomial of degree k. Note that in this case Euler summation reduces an infinite series to a finite sum.
• The particular choice $P_k(j):= (j+1)^k$ provides an explicit representation of the Bernoulli numbers, since $\zeta(-k)= -\frac{B_{k+1}}{k+1}$. Indeed, applying Euler summation to the zeta function yields $\frac{1}{1-2^{k+1}}\sum_{i=0}^k \frac{1}{2^{i+1}} \sum_{j=0}^i {i \choose j} (-1)^j (j+1)^k$, which is polynomial for $k$ a positive integer; cf. Riemann zeta function.
• $\sum_{j=0}^\infty z^j= \sum_{i=0}^\infty \frac{1}{(1+y)^{i+1}} \sum_{j=0}^i {i \choose j} y^{j+1} z^j = \frac{y}{1+y} \sum_{i=0} \left( \frac{1+yz}{1+y} \right)^i$. With an appropriate choice of $y$ this series converges to $\frac{1}{1-z}$.