Approximaths

Month: November 2025

  • ‘Generic’ summation II

    Consider the series:

    \displaystyle 1+0-1+1+0-1+1+0-1+\dots

    using the generic summation (summation “machine”):

    \displaystyle \begin{aligned} s &= \mathcal{S}(1+0-1+1+0-1+1+0-1+\dots) \\ s &= 1 + \mathcal{S}(0-1+1+0-1+1+0-1+\dots) \\ s &= 1 + \mathcal{S}(-1+1+0-1+1+0-1+\dots) \end{aligned}

    Sum these 3 equations term by term:

    \displaystyle \begin{aligned} 3s &= 2 + \mathcal{S}(0+0+0+0+0+0+0+\dots) \\ s &= \frac{2}{3} \end{aligned}

    Now apply Euler summation to the same series:

    \displaystyle 1+0-1+1+0-1+1+0-1+\dots

    We multiply each n-term with x^n:

    \displaystyle \begin{aligned} f(x) &:= 1x^0+0x^1-1x^2+1x^3+0x^4-1x^5+1x^6+0x^7-1x^8+\dots \\ &:= 1-x^2+x^3-x^5+x^6-x^8+\dots \\ &:= (1+x^3+x^6+x^9+\dots) - (x^2+x^5+x^8+\dots) \end{aligned}

    In this last step we changed the order of summation of a Taylor series. Inside its radius of convergence this step is valid. We have:

    \displaystyle \begin{aligned} f(x) &= (1+x^3+x^6+x^9+\dots) - (x^2+x^5+x^8+\dots) \\ &= (1+x^3+x^6+x^9+\dots) - x^2(1+x^3+x^6+\dots) \\ &= ((x^3)^0+(x^3)^1+(x^3)^2+(x^3)^3+\dots) - x^2((x^3)^0+(x^3)^1+(x^3)^2+(x^3)^3+\dots) \\ &= \frac{1}{1-x^3} - x^2\frac{1}{1-x^3} \\ &= \frac{1-x^2}{1-x^3} \end{aligned}

    Now calculate the limit (using Hospital’s rule):

    \displaystyle \lim_{x\to 1} \frac{1-x^2}{1-x^3} = \frac{2}{3}

    The generic summation “machine” and the Euler summation deliver the same result.

  • ‘Generic’ summation I

    Let imagine that we have engineered a summation ‘machine’ called \mathcal{S}() and consider the geometric series:

    \displaystyle 1+x+x^2+x^3+x^4+\dots

    Apply our summation ‘machine’ on this series and call the result s:

    \displaystyle s = \mathcal{S}(1+x+x^2+x^3+x^4+\dots)

    We assign the following two properties for the ‘machine’ \mathcal{S}:

    \displaystyle \mathcal{S}(a_0+a_1+a_2+\dots) = a_0 + \mathcal{S}(a_1+a_2+\dots) \quad \text{(first property)}
    \displaystyle \mathcal{S}(\sum(\alpha a_n) + \sum(\beta b_n)) = \alpha \mathcal{S}(\sum a_n) + \beta \mathcal{S}(\sum b_n)) \quad \text{(second property)}

    where \alpha and \beta are constants. Equipped with the machine \mathcal{S}() and its two properties consider again the geometric series:

    \displaystyle \begin{aligned} s &= \mathcal{S}(1+x+x^2+x^3+x^4+\dots) &\quad\text{(definition)}\\ s &= 1+ \mathcal{S}(x+x^2+x^3+x^4+\dots) &\quad\text{(first property)}\\ s &= 1+ x\mathcal{S}(1+x+x^2+x^3+\dots) &\quad\text{(second property)}\\ s &= 1+ xs &\quad\text{(definition)}\\ s &= \frac{1}{1-x} \end{aligned}

    Consider the series:

    \displaystyle 1-1+1-1+1-1+\dots

    If we use traditional, ‘rigorous’, summation techniques we will conclude that this alternating series is not converging. Now apply the summation “machine” as described above:

    \displaystyle \begin{aligned} s &= \mathcal{S}(1-1+1-1+1-1+\dots ) \\ s &= 1+ \mathcal{S}(-1+1-1+1-1+\dots ) \\ s &= 1 -\mathcal{S}(1-1+1-1+1-\dots ) \\ s &= 1-s \\ s &= \frac{1}{2} \end{aligned}

  • Borel summation

    Let introduce the Borel summation by first recalling the following property:

    \displaystyle n!= \int_{0}^{\infty} e^{-t}t^{n}dt
    \displaystyle 1 = \frac{\int_{0}^{\infty} e^{-t}t^{n}dt}{n!}

    We would like to sum the following series:

    \displaystyle \sum_{n=0}^{\infty}a_n = a_0 + a_1 + a_2 + a_3 + a_4 +...
    \displaystyle \phantom{\sum_{n=0}^{\infty}a_n} = 1a_0 + 1a_1 + 1a_2 + 1a_3 + 1a_4 +...
    \displaystyle \phantom{\sum_{n=0}^{\infty}a_n} = \sum_{n=0}^{\infty} \frac{\int_{0}^{\infty} e^{-t}t^{n}dt}{n!} a_n

    The Borel sum B is defined as follow:

    \displaystyle B := \sum_{n=0}^{\infty} \frac{\int_{0}^{\infty} e^{-t}t^{n}dt}{n!} a_n
    \displaystyle \phantom{B :=} = \int_{0}^{\infty}e^{-t} \sum_{n=0}^{\infty} \frac{t^n}{n!}a_n\,dt

    The factorial term makes this sum with better chance to converge. In the section concerning the Euler summation we have seen that:

    \displaystyle E(1 - 1 + 1 - 1 + 1 - 1 + ...) = \frac{1}{2}

    Let calculate the corresponding Borel sum:

    \displaystyle B(1 - 1 + 1 - 1 + 1 - 1 + ...) = \int_{0}^{\infty}e^{-t} \sum_{n=0}^{\infty} \frac{t^n}{n!}(-1)^n \,dt
    \displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = \int_{0}^{\infty}e^{-t} \sum_{n=0}^{\infty} \frac{(-t)^n}{n!} \,dt
    \displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = \int_{0}^{\infty} e^{-t}e^{-t}\,dt
    \displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = \int_{0}^{\infty} e^{-2t}\,dt
    \displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = -\frac{1}{2} e^{-2t}\Big|_{0}^{\infty}
    \displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = 0 - (-\frac{1}{2})
    \displaystyle \phantom{B(1 - 1 + 1 - 1 + 1 - 1 + ...)} = \frac{1}{2}

    and therefore:

    \displaystyle B(1-1+1-1+1-1+...) = E(1-1+1-1+1-1+...)

    Consider the series:

    \displaystyle -1+2!-3!+4!-5!+6!-...
    \displaystyle B(-1+2!-3!+4!-...) = \int_{0}^{\infty}e^{-t} \sum_{n=1}^{\infty} \frac{t^n}{n!}(-1)^n n! \,dt
    \displaystyle \phantom{B(-1+2!-3!+4!-...)} = \int_{0}^{\infty}e^{-t} \sum_{n=1}^{\infty} (-t)^n \,dt

    \sum_{n=0}^{\infty} (-t)^n converges for t < 1 to -\frac{1}{1+t} therefore:

    \displaystyle B = -\int_{0}^{\infty} \frac{te^{-t}}{(1+t)} \,dt \approx -0.40365

    If a series is summable in the sense of Euler, then it is also summable in the sense of Borel, and both summation methods yield the same value. The converse is false: there exist series that are summable in the sense of Borel but not in the sense of Euler. In other words, Borel summation is more powerful, as it applies to more strongly divergent series.

  • Euler summation II

    In the previous post, we introduced Euler summation. The following are two examples where it fails to produce a finite result.

    Consider the divergent series:

    \displaystyle 0 + 1 + 2 + 3 + 4 + \dots

    Define:

    \displaystyle \begin{array}{rcl} f(x) &=& 0x^{0} + 1x^{1} + 2x^{2} + 3x^{3} + 4x^{4} + \dots \\ &=& \sum_{n=0}^{\infty} nx^{n} = \frac{x}{(1-x)^{2}} \\ \end{array}

    The Euler sum is:

    \displaystyle E = \lim_{x \to 1_{-}} \frac{x}{(1-x)^{2}} = \infty
    \displaystyle E(0+ 1 + 2 + 3 + 4 + \dots) = \infty

    Now consider the divergent series:

    \displaystyle 1 + 4 + 9 + 16 + 25 + 36 + \dots

    Define:

    \displaystyle \begin{array}{rcl} f(x) &=& 1^{2}x^{1}+2^{2}x^{2}+3^{2}x^{3}+4^{2}x^{4}+5^{2}x^{5}+6^{2}x^{6}+\dots \\ &=& \sum_{n=1}^{\infty} n^{2} x^{n} = \frac{x(1+x)}{(1-x)^{3}} \\ \end{array}

    The Euler sum is:

    \displaystyle E = \lim_{x \to 1_{-}}\frac{x(1+x)}{(1-x)^{3}} = \infty
    \displaystyle E(1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots) = \infty

    Advantages

    Regularization of slowly divergent series:
    Euler summation can assign a finite value to some divergent series that oscillate or diverge slowly, such as

    \displaystyle 1 - 1 + 1 - 1 + \dots

    where E(series) = \tfrac{1}{2}.

    Improved convergence:
    For many convergent series, Euler transformation accelerates convergence, making it useful for numerical computations.

    Analytic continuation link:
    It provides a bridge between ordinary summation and more advanced summation methods (e.g. Borel or zeta regularization).

    Disadvantages

    Limited domain of applicability:
    Euler summation fails for series that diverge too rapidly, such as

    \displaystyle 1 + 2 + 3 + 4 + \dots

    where E(series) = \infty.

    Not uniquely defined for all divergent series:
    Some series cannot be assigned a finite Euler sum, or the method may yield inconsistent results depending on the transformation order.

    Weaker than analytic regularization:
    Compared to zeta or Borel summation, Euler’s method handles fewer classes of divergent series and lacks a rigorous analytic continuation framework.